Codeforces Round #450 (Div. 2) ABCD

这次还是能看的0 0，没出现一题掉分情况。

QAQ前两次掉分还被hack了0 0,两行清泪。

A. Find Extra One

 

You have n distinct points on a plane, none of them lie on OY axis. Check that there is a point after removal of which the remaining points are located on one side of the OY axis.

Input

The first line contains a single positive integer n (2 ≤ n ≤ 105).

The following n lines contain coordinates of the points. The i-th of these lines contains two single integers xi and yi (|xi|, |yi| ≤ 109,xi ≠ 0). No two points coincide.

Output

Print "Yes" if there is such a point, "No" — otherwise.

You can print every letter in any case (upper or lower).

Examples

input

3
1 1
-1 -1
2 -1

output

Yes

input

4
1 1
2 2
-1 1
-2 2

output

No

input

3
1 2
2 1
4 60

output

Yes

---

题意：给你n个平面上的点，能否去掉一个使得所有点在y轴一侧。给出的点不会在y轴上。

题解：水题，统计下x>0 和 x<0的数字个数就好了。

#include<bits/stdc++.h>
#define clr(x) memset(x,0,sizeof(x))
#define clr_1(x) memset(x,-1,sizeof(x))
#define mod 1000000007
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;
int main()
{
    int n,left,right,x,y;
    left=right=0;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d",&x,&y);
        if(x>0)
            right++;
        else
            left++;
    }
    if(right<=1 || left<=1)
        printf("Yes
");
    else
        printf("No
");
    return 0;
}

B. Position in Fraction

 

You have a fraction . You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.

Input

The first contains three single positive integers a, b, c (1 ≤ a < b ≤ 105, 0 ≤ c ≤ 9).

Output

Print position of the first occurrence of digit c into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.

Examples

input

1 2 0

output

2

input

2 3 7

output

-1

---

题意：给出一个分数$ frac{a}{b} $ ，看c在该分数小数点后几位。

题解：$ frac{a}{b} $化为最简$ frac{a'}{b'} $后，我们模拟除法列竖式求答案的过程。由于列竖式的过程中余下的数总比b'小，所以最多b'个不同余数，列竖式的过程中一旦出现相同的余数则是进入了循环。所以我们最多做b'次除法就能求出c是否存在于该分数小数点后以及c的位置。

#include<bits/stdc++.h>
#define clr(x) memset(x,0,sizeof(x))
#define clr_1(x) memset(x,-1,sizeof(x))
#define mod 1000000007
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;
int gcd(int a,int b)
{
    int c;
    while(b)
    {
        c=a%b;
        a=b;
        b=c;
    }
    return a;
}
int main()
{
    int a,b,c,d,e,i;
    scanf("%d%d%d",&a,&b,&c);
    d=gcd(a,b);
    a/=d;
    b/=d;
    d=a/b;
    a-=d*b;
    a*=10;
    for(i=1;i<=100000;i++)
    {
        d=a/b;
        a-=d*b;
        if(d==c)
            break;
        a*=10;
    }
    if(i>100000)
        printf("-1
");
    else
        printf("%d
",i);
    return 0;
}

C. Remove Extra One

 

You are given a permutation p of length n. Remove one element from permutation to make the number of records the maximum possible.

We remind that in a sequence of numbers a1, a2, ..., ak the element ai is a record if for every integer j (1 ≤ j < i) the following holds:aj < ai.

Input

The first line contains the only integer n (1 ≤ n ≤ 105) — the length of the permutation.

The second line contains n integers p1, p2, ..., pn (1 ≤ pi ≤ n) — the permutation. All the integers are distinct.

Output

Print the only integer — the element that should be removed to make the number of records the maximum possible. If there are multiple such elements, print the smallest one.

Examples

input

1
1

output

1

input

5
5 1 2 3 4

output

5

Note

In the first example the only element can be removed.

---

题意:给出1~n的一个排列，要求你找出一个数，在排列中去掉它以后符合条件的数最多，如果有多个则选最小的那个。若ai符合条件，对于任意1≤j<i，aj<a_i。

题解：那么我们写个bit来记录数列前面小于ai的个数，set来求取大于ai的第一个数。如果ai前面有i-2个数小于ai，那么用set的lonwer_bound找到大于a_i的第一个数p，并且该数p对于答案的贡献num[p]++。注意如果ai前面有i-1个数小于a_i，那么a_i对于答案的贡献num[a_i]--。然后从小到大找贡献最大的那个。

#include<bits/stdc++.h>
#define clr(x) memset(x,0,sizeof(x))
#define clr_1(x) memset(x,-1,sizeof(x))
#define mod 1000000007
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;
const int N=1e5+10;
int bits[N];
int num[N];
int n,m,T,k,p;
set<int> pt;
set<int>::iterator it;
void add(int i,int x)
{
    while(i<=n)
    {
        bits[i]+=x;
        i+=(i &(-i));
    }
    return ;
}
int sum(int i)
{
    int res=0;
    while(i>0)
    {
        res+=bits[i];
        i-=(i &(-i));
    }
    return res;
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&p);
        if((p-1>0 && sum(p-1)==i-2) || (p-1==0 && i==2))
        {
            it=pt.lower_bound(p);
            num[*it]++;
        }
        if((p-1>0 && sum(p-1)==i-1) || (p-1==0 && i==1))
            num[p]--;
        pt.insert(p);
        add(p,1);
    }
    int maxn=num[1];
    k=1;
    for(int i=2;i<=n;i++)
    {
        if(maxn<num[i])
        {
            maxn=num[i];
            k=i;
        }
    }
    printf("%d
",k);
    return 0;
}

D. Unusual Sequences

 

Count the number of distinct sequences a1, a2, ..., an (1 ≤ ai) consisting of positive integers such that gcd(a1, a2, ..., an) = x and . As this number could be large, print the answer modulo 109 + 7.

gcd here means the greatest common divisor.

Input

The only line contains two positive integers x and y (1 ≤ x, y ≤ 109).

Output

Print the number of such sequences modulo 109 + 7.

Examples

input

3 9

output

3

input

5 8

output

0

---

题意：让你找出符合条件的序列数，并mod 1e9+7。序列需满足gcd（a_1~n）=x,sum(a_1~n）=y。

题解：首先把一个数p分解成几个数成组成的序列有$ 2^{p-1} $，用组合的隔板法轻易可证。x能整除y说明存在这样的序列，反之不存在。然后d=y/x，将d分解成质数乘积，算算不同质数的个数以及具体是哪些质数，容斥一下就能求出答案了。

#include<bits/stdc++.h>
#define clr(x) memset(x,0,sizeof(x))
#define clr_1(x) memset(x,-1,sizeof(x))
#define mod 1000000007
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;
const int N=1e5+10;
int prime[N],inf[N],pcnt;
void primer(int n)
{
    pcnt=0;
    for(int i=2;i<=n;i++)
    {
        if(!inf[i])
        {
            prime[++pcnt]=i;
        }
        for(int j=1;j<=pcnt;j++)
        {
            if(prime[j]>n/i) break;
            inf[prime[j]*i]=1;
            if(i%prime[j]==0) break;
        }
    }
    return ;
}
LL quick_pow(LL x,LL n)
{
    LL res=1;
    while(n)
    {
        if(n&1)
            res=(res*x)%mod;
        n>>=1;
        x=(x*x)%mod;
    }
    return res;
}
LL  x,y,d,n,k;
int all;
LL num[20];
int cnt;
LL ans,p;
int main()
{
    primer(100000);
    scanf("%lld%lld",&x,&y);
    if(y%x!=0)
    {
        printf("0
");
        return 0;
    }
    y/=x;
    d=y;
    cnt=0;
    for(int i=1;(LL)prime[i]*prime[i]<=d;i++)
    {
        n=(LL)prime[i];
        if(d%n==0)
        {
            num[++cnt]=n;
            while(d%n==0) d/=n;
        }
    }
    if(d!=1) num[++cnt]=d;
    ans=0;
    all=(1<<cnt)-1;
    for(int i=0,j,k,ok;i<=all;i++)
    {
        j=i;
        k=0;
        p=1;
        ok=1;
        while(j)
        {
            k++;
            if(j&1) p*=num[k],ok=-ok;
            j>>=1;
        }
        ans=(ans+quick_pow(2,y/p-1)*ok)%mod;
    }
    printf("%lld
",(ans%mod+mod)%mod);
    return 0;
}