2019寒假培训第二天

A - Diverse Team

There are n students in a school class, the rating of the i-th student on Codehorses is ai. You have to form a team consisting of k students (1≤k≤n

) such that the ratings of all team members are distinct.

If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print k

distinct numbers which should be the indices of students in the team you form. If there are multiple answers, print any of them.

Input

The first line contains two integers n

and k (1≤k≤n≤100

) — the number of students and the size of the team you have to form.

The second line contains n

integers a1,a2,…,an (1≤ai≤100), where ai is the rating of i

-th student.

Output

If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print k

distinct integers from 1 to n

which should be the indices of students in the team you form. All the ratings of the students in the team should be distinct. You may print the indices in any order. If there are multiple answers, print any of them.

Assume that the students are numbered from 1

to n

.

Examples

Input

5 3
15 13 15 15 12

Output

YES
1 2 5 

Input

5 4
15 13 15 15 12

Output

NO

Input

4 4
20 10 40 30

Output

YES
1 2 3 4 

Note

All possible answers for the first example:

• {1 2 5}
• {2 3 5}
• {2 4 5}

Note that the order does not matter.

代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,k,z,m=0,a[105],b[105];
    bool flag;
    
    cin>>n>>k;
    for(int i=1; i<=n; ++i)
    {
        cin>>z;
        flag=false;
        for(int j=0; j<m; ++j)
            if(a[j]==z)
            {
                flag=true;
                break;
            }
        if(!flag)
        {
            b[m]=i;
            a[m++]=z;
        }
    }
    if(m>=k)
    {
        cout<<"YES"<<endl;
        for(int i=0; i<k-1; ++i)
            cout<<b[i]<<" ";
        cout<<b[k-1]<<endl;
    }
    else cout<<"NO"<<endl;
    return 0;
}

G - Remove Duplicates

Petya has an array a consisting of n

integers. He wants to remove duplicate (equal) elements.

Petya wants to leave only the rightmost entry (occurrence) for each element of the array. The relative order of the remaining unique elements should not be changed.

Input

The first line contains a single integer n

(1≤n≤50

) — the number of elements in Petya's array.

The following line contains a sequence a1,a2,…,an

(1≤ai≤1000

) — the Petya's array.

Output

In the first line print integer x

— the number of elements which will be left in Petya's array after he removed the duplicates.

In the second line print x

integers separated with a space — Petya's array after he removed the duplicates. For each unique element only the rightmost entry should be left.

Examples

Input

6
1 5 5 1 6 1

Output

3
5 6 1 

Input

5
2 4 2 4 4

Output

2
2 4 

Input

5
6 6 6 6 6

Output

1
6 

Note

In the first example you should remove two integers 1

, which are in the positions 1 and 4. Also you should remove the integer 5, which is in the position 2

.

In the second example you should remove integer 2

, which is in the position 1, and two integers 4, which are in the positions 2 and 4

.

In the third example you should remove four integers 6

, which are in the positions 1, 2, 3 and 4.

 代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    int v[1100]={0};
    int a[1100],b[1100];
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
    }
    int j=0;
    for(int i=n-1;i>=0;i--)
    {
        if(v[a[i]]==0)
        {
            v[a[i]]=1;
            b[j++]=a[i];
        }
    }
    cout<<j<<endl;
    for(int k=j-1;k>=1;k--)
    {
        cout<<b[k]<<" ";
    }
    cout<<b[0]<<endl;
    return 0;
}

B - Substrings Sort

You are given n

strings. Each string consists of lowercase English letters. Rearrange (reorder) the given strings in such a way that for every string, all strings that are placed before it are its substrings.

String a

is a substring of string b if it is possible to choose several consecutive letters in b in such a way that they form a

. For example, string "for" is contained as a substring in strings "codeforces", "for" and "therefore", but is not contained as a substring in strings "four", "fofo" and "rof".

Input

The first line contains an integer n

(1≤n≤100

) — the number of strings.

The next n

lines contain the given strings. The number of letters in each string is from 1 to 100

, inclusive. Each string consists of lowercase English letters.

Some strings might be equal.

Output

If it is impossible to reorder n

given strings in required order, print "NO" (without quotes).

Otherwise print "YES" (without quotes) and n

given strings in required order.

Examples

Input

5
a
aba
abacaba
ba
aba

Output

YES
a
ba
aba
aba
abacaba

Input

5
a
abacaba
ba
aba
abab

Output

NO

Input

3
qwerty
qwerty
qwerty

Output

YES
qwerty
qwerty
qwerty

Note

In the second example you cannot reorder the strings because the string "abab" is not a substring of the string "abacaba".

代码：

#include<bits/stdc++.h>
using namespace std;
bool cmp(string a,string b)
{
    return a.size()<b.size();
}
int main()
{
    int n;
    cin>>n;
    string s[110];
    for(int i=0; i<n; i++)
    {
        cin>>s[i];
    }
    sort(s,s+n,cmp);
    bool flag=true;
    for(int i=0; i<n-1; i++)
    {
        if(s[i+1].find(s[i])==string::npos)//找不到
            flag=false;
    }
    if(!flag)
        cout<<"NO"<<endl;
    else
    {
        cout<<"YES"<<endl;
        for(int i=0; i<n; i++)
        {

            cout<<s[i]<<endl;
        }
    }
    return 0;
}

I - Letters

• There are n dormitories in Berland State University, they are numbered with integers from 1 to n. Each dormitory consists of rooms, there are ai rooms in i-th dormitory. The rooms in i-th dormitory are numbered from 1 to ai

  .

  A postman delivers letters. Sometimes there is no specific dormitory and room number in it on an envelope. Instead of it only a room number among all rooms of all n

  dormitories is written on an envelope. In this case, assume that all the rooms are numbered from 1 to a1+a2+⋯+an

  and the rooms of the first dormitory go first, the rooms of the second dormitory go after them and so on.

  For example, in case n=2

  , a1=3 and a2=5 an envelope can have any integer from 1 to 8 written on it. If the number 7 is written on an envelope, it means that the letter should be delivered to the room number 4

  of the second dormitory.

  For each of m

  letters by the room number among all n

  dormitories, determine the particular dormitory and the room number in a dormitory where this letter should be delivered.

  Input

  The first line contains two integers n

  and m (1≤n,m≤2⋅105)

  — the number of dormitories and the number of letters.

  The second line contains a sequence a1,a2,…,an

  (1≤ai≤1010), where ai equals to the number of rooms in the i-th dormitory. The third line contains a sequence b1,b2,…,bm (1≤bj≤a1+a2+⋯+an), where bj equals to the room number (among all rooms of all dormitories) for the j-th letter. All bj

  are given in increasing order.

  Output

  Print m

  lines. For each letter print two integers f and k — the dormitory number f (1≤f≤n) and the room number k in this dormitory (1≤k≤af)

  to deliver the letter.

  Examples
  Input

  3 6
  10 15 12
  1 9 12 23 26 37

  Output

  1 1
  1 9
  2 2
  2 13
  3 1
  3 12

  Input

  2 3
  5 10000000000
  5 6 9999999999

  Output

  1 5
  2 1
  2 9999999994

  Note

  In the first example letters should be delivered in the following order:

  • the first letter in room 1
• of the first dormitory
• the second letter in room 9
• of the first dormitory
• the third letter in room 2
• of the second dormitory
• the fourth letter in room 13
• of the second dormitory
• the fifth letter in room 1
• of the third dormitory
• the sixth letter in room 12
of the third dormitory

知识点：前缀和+lower_bound(begin,end,b)

代码：

#include<bits/stdc++.h>
using namespace std;
#define MAXN 200010
typedef long long ll;
    int n,m;
    ll a[MAXN],b;
    ll c[MAXN];
int main()
{

    cin>>n>>m;
    memset(c,0,sizeof c);
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        c[i]=c[i-1]+a[i];
    }
    for(int i=1;i<=m;i++)
    {
         cin>>b;
         int pos=lower_bound(c+1,c+1+n,b)-(c);
         cout<<pos<<" "<<b-c[pos-1]<<endl;
    }
    return 0;
}