Educational Codeforces Round 42 (Rated for Div. 2) A

A. Equator

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarp has created his own training plan to prepare for the programming contests. He will train for nn days, all days are numbered from 11to nn, beginning from the first.

On the ii-th day Polycarp will necessarily solve aiai problems. One evening Polycarp plans to celebrate the equator. He will celebrate it on the first evening of such a day that from the beginning of the training and to this day inclusive he will solve half or more of all the problems.

Determine the index of day when Polycarp will celebrate the equator.

Input

The first line contains a single integer nn (1≤n≤2000001≤n≤200000) — the number of days to prepare for the programming contests.

The second line contains a sequence a1,a2,…,ana1,a2,…,an (1≤ai≤100001≤ai≤10000), where aiai equals to the number of problems, which Polycarp will solve on the ii-th day.

Output

Print the index of the day when Polycarp will celebrate the equator.

Examples

input

Copy

4
1 3 2 1

output

Copy

2

input

Copy

6
2 2 2 2 2 2

output

Copy

3

Note

In the first example Polycarp will celebrate the equator on the evening of the second day, because up to this day (inclusive) he will solve 44out of 77 scheduled problems on four days of the training.

In the second example Polycarp will celebrate the equator on the evening of the third day, because up to this day (inclusive) he will solve 66out of 1212 scheduled problems on six days of the training.

 题意：求一组数据中大于等于和的一半的数的位置在哪

tips：一开始我用的是找的 t>sum/2的位置，然后被hack了，冷静分析了一波

如果输入的数据是      6     1 2 3 1 2 4时

得到的答案是 

但是实际上应该是4，因为 ‘/’ 这个是向0取整，所sum/2=6，但是在i=3这个地方时 t=6，所以为了‘/’所产生的误差 就用t*2>sum来判断 就不会产生误差

实际上是一个水题，但是要注意细节

附上代码

#include<bits/stdc++.h>
using namespace std;
const int maxx=;
typedef long long ll;
int a[maxx];
int main()
{
    int n;
    ll s=;
    scanf("%d",&n);
    for(int i=;i<=n;i++){
        scanf("%d",&a[i]);
        s+=a[i];
    }
    ll t=;
    for(int i=;i<=n;i++){
        t+=a[i];
        if(t*>=s){
            printf("%d\n",i);
            break;
        }
    }
    return ;
}