Codeforces Round #371 (Div. 2) A ，B ， C 水，水，trie树

A. Meeting of Old Friends

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya!

Sonya is an owl and she sleeps during the day and stay awake from minute l1 to minute r1 inclusive. Also, during the minute k she prinks and is unavailable for Filya.

Filya works a lot and he plans to visit Sonya from minute l2 to minute r2 inclusive.

Calculate the number of minutes they will be able to spend together.

Input

The only line of the input contains integers l1, r1, l2, r2 and k (1 ≤ l1, r1, l2, r2, k ≤ 1018, l1 ≤ r1, l2 ≤ r2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.

Output

Print one integer — the number of minutes Sonya and Filya will be able to spend together.

Examples

input

1 10 9 20 1

output

2

input

1 100 50 200 75

output

50

Note

In the first sample, they will be together during minutes 9 and 10.

In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100.

题意：两个区间交集，去掉k这个点；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
const int N=1e5+10,M=1e6+1010,inf=1e9+10,mod=1e9+7;
const ll INF=1e18+10;
int main()
{
    ll l1,r1,l2,r2,k;
    scanf("%lld%lld%lld%lld%lld",&l1,&r1,&l2,&r2,&k);
    ll l=max(l1,l2);
    ll r=min(r1,r2);
    if(l>r)
    printf("0
");
    else if(k>=l&&k<=r)
    printf("%lld
",max(r-l,0LL));
    else
    printf("%lld
",max(0LL,r-l+1));
    return 0;
}

B. Filya and Homework

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Today, hedgehog Filya went to school for the very first time! Teacher gave him a homework which Filya was unable to complete without your help.

Filya is given an array of non-negative integers a1, a2, ..., an. First, he pick an integer x and then he adds x to some elements of the array (no more than once), subtract x from some other elements (also, no more than once) and do no change other elements. He wants all elements of the array to be equal.

Now he wonders if it's possible to pick such integer x and change some elements of the array using this x in order to make all elements equal.

Input

The first line of the input contains an integer n (1 ≤ n ≤ 100 000) — the number of integers in the Filya's array. The second line containsn integers a1, a2, ..., an (0 ≤ ai ≤ 109) — elements of the array.

Output

If it's impossible to make all elements of the array equal using the process given in the problem statement, then print "NO" (without quotes) in the only line of the output. Otherwise print "YES" (without quotes).

Examples

input

5
1 3 3 2 1

output

YES

input

5
1 2 3 4 5

output

NO

Note

In the first sample Filya should select x = 1, then add it to the first and the last elements of the array and subtract from the second and the third elements.

题意：找到一个x，将数组中每个数+x，-x，不变使得数组全为一个数；

思路：去重，剩下的数个数大于3肯定no，小于3肯定yes，等于3，判断是否等差；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
const int N=1e5+10,M=1e6+1010,inf=1e9+10,mod=1e9+7;
const ll INF=1e18+10;
int a[N];
int main()
{
    int x;
    scanf("%d",&x);
    for(int i=0;i<x;i++)
    scanf("%d",&a[i]);
    sort(a,a+x);
    int ans=unique(a,a+x)-a;
    if(ans>3)
    printf("NO
");
    else if(ans<3)
    printf("YES
");
    else
    {
        sort(a,a+3);
        if(a[1]-a[0]==a[2]-a[1])
        printf("YES
");
        else
        printf("NO
");
    }
    return 0;
}

C. Sonya and Queries

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Today Sonya learned about long integers and invited all her friends to share the fun. Sonya has an initially empty multiset with integers. Friends give her t queries, each of one of the following type:

1.  +  ai — add non-negative integer ai to the multiset. Note, that she has a multiset, thus there may be many occurrences of the same integer.
2.  -  ai — delete a single occurrence of non-negative integer ai from the multiset. It's guaranteed, that there is at least one ai in the multiset.
3. ? s — count the number of integers in the multiset (with repetitions) that match some pattern s consisting of 0 and 1. In the pattern, 0stands for the even digits, while 1 stands for the odd. Integer x matches the pattern s, if the parity of the i-th from the right digit in decimal notation matches the i-th from the right digit of the pattern. If the pattern is shorter than this integer, it's supplemented with0-s from the left. Similarly, if the integer is shorter than the pattern its decimal notation is supplemented with the 0-s from the left.

For example, if the pattern is s = 010, than integers 92, 2212, 50 and 414 match the pattern, while integers 3, 110, 25 and 1030 do not.

Input

The first line of the input contains an integer t (1 ≤ t ≤ 100 000) — the number of operation Sonya has to perform.

Next t lines provide the descriptions of the queries in order they appear in the input file. The i-th row starts with a character ci — the type of the corresponding operation. If ci is equal to '+' or '-' then it's followed by a space and an integer ai (0 ≤ ai < 1018) given without leading zeroes (unless it's 0). If ci equals '?' then it's followed by a space and a sequence of zeroes and onse, giving the pattern of length no more than 18.

It's guaranteed that there will be at least one query of type '?'.

It's guaranteed that any time some integer is removed from the multiset, there will be at least one occurrence of this integer in it.

Output

For each query of the third type print the number of integers matching the given pattern. Each integer is counted as many times, as it appears in the multiset at this moment of time.

Examples

input

12
+ 1
+ 241
? 1
+ 361
- 241
? 0101
+ 101
? 101
- 101
? 101
+ 4000
? 0

output

2
1
2
1
1

input

4
+ 200
+ 200
- 200
? 0

output

1

Note

Consider the integers matching the patterns from the queries of the third type. Queries are numbered in the order they appear in the input.

1. 1 and 241.
2. 361.
3. 101 and 361.
4. 361.
5. 4000.

 思路：trie数模板题，map应该也可以过；

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define mod 1000000007
#define pi (4*atan(1.0))
const int N=1e5+10,M=4e6+10,inf=1e9+10;
int a[M][3],sum[M],len;
void init()
{
    memset(a,0,sizeof(a));
    memset(sum,0,sizeof(sum));
    len=1;
}
void insertt(ll x)
{
    int num[20];
    memset(num,0,sizeof(num));
    int flag=0;
    while(x)
    {
        num[flag++]=x%2;
        x/=10;
    }
    int u=0,n=19;
    for(int i=n; i>=0; i--)
    {
        if(!a[u][num[i]])
        {
            a[u][num[i]]=len++;
        }
        u=a[u][num[i]];
        sum[u]++;
    }
}
void del(ll x)
{
    int num[20];
    memset(num,0,sizeof(num));
    int flag=0;
    while(x)
    {
        num[flag++]=x%2;
        x/=10;
    }
    int u=0,n=19;
    for(int i=n; i>=0; i--)
    {
        int v=a[u][num[i]];
        sum[v]--;
        if(!sum[v])
        a[u][num[i]]=0;
        u=v;
    }
}
char gg[20];
int getans(ll x)
{
    scanf("%s",&gg);
    int num[20];
    memset(num,0,sizeof(num));
    int flag=0;
    for(int i=strlen(gg)-1;i>=0;i--)
    num[flag++]=gg[i]-'0';
    int u=0,n=19,v,w;
    int ans=0;
    for(int i=n; i>=0; i--)
    {
        if(!a[u][num[i]])
            return 0;
        u=a[u][num[i]];
    }
    return sum[u];
}
char ch[10];
int main()
{
    int T,cas;
    ll x;
    init();
    scanf("%d",&T);
    for(cas=1; cas<=T; cas++)
    {
        scanf("%s%",ch);
        if(ch[0]=='+')
        {
            scanf("%lld",&x);
            insertt(x);
        }
        else if(ch[0]=='-')
        {
            scanf("%lld",&x);
            del(x);
        }
        else
        {
            printf("%d
",getans(x));
        }

    }
    return 0;
}