Codeforces Round #450 (Div. 2)

这次补的特别顺利，因为在群里看到他们结论了啊

A. Find Extra One

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 xiand 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

Note

In the first example the second point can be removed.

In the second example there is no suitable for the condition point.

In the third example any point can be removed.

 

 上题意就行，这个题目考英语，题目很长

#include<bits/stdc++.h>
using namespace std;
int t1,t2,n;
int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        if(x>0)t1++;
        else if(x<0)t2++;
    }
    if(t1<=1||t2<=1)
        printf("Yes\n");
    else
        printf("No\n");
    return 0;    
} 

B. Position in Fraction

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

Note

The fraction in the first example has the following decimal notation: . The first zero stands on second position.

The fraction in the second example has the following decimal notation: . There is no digit 7 in decimal notation of the fraction.

 

 这个B还是很有意思的，问你a/b最先出现的c在哪一位，一般人都不会想到去暴力吧，那个无限不循环小数恶心到爆炸，但是庆幸的是a,b的大小，a/b最大也是1e5的,循环节是不会超过1e6的

#include<bits/stdc++.h>
using namespace std;
int a,b,c,x;
int main()
{
    scanf("%d%d%d",&a,&b,&c);
    for(int i=1;i<=1e6;i++)
    {
        a*=10;
        if(a>=b)
        {
            x=a/b;
            a=a%b;
        }
        else
            x=0;
        if(x==c)
        {
            printf("%d\n",i);
            return 0;
        }
    }
    printf("-1\n");
    return 0;
}

 

C. Remove Extra One

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

 

 这个题目可以这样转换一下啊

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int mx,mx2,n,mn=-1e9,res;
int a[N];
int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        int x;
        scanf("%d",&x);
        if(x>mx)
        {
            mx2=mx;
            mx=max(mx,x);
            a[x]--;
        }
        else if(x>mx2)
        {
            a[mx]++;
            mx2=max(mx2,x);
        }
    }
    for(int i=1;i<=n;i++)
    {
        if(a[i]>mn)
        {
            mn=a[i];
            res=i;
        }
    }
    cout<<res<<endl;
    return 0;
}