C

Problem description

Alice and Bob like games. And now they are ready to start a new game. They have placed n chocolate bars in a line. Alice starts to eat chocolate bars one by one from left to right, and Bob — from right to left. For each chocololate bar the time, needed for the player to consume it, is known (Alice and Bob eat them with equal speed). When the player consumes a chocolate bar, he immediately starts with another. It is not allowed to eat two chocolate bars at the same time, to leave the bar unfinished and to make pauses. If both players start to eat the same bar simultaneously, Bob leaves it to Alice as a true gentleman.

How many bars each of the players will consume?

Input

The first line contains one integer n (1 ≤ n ≤ 10⁵) — the amount of bars on the table. The second line contains a sequence t₁, t₂, ..., t_n (1 ≤ t_i ≤ 1000), where t_i is the time (in seconds) needed to consume the i-th bar (in the order from left to right).

Output

Print two numbers a and b, where a is the amount of bars consumed by Alice, and b is the amount of bars consumed by Bob.

Examples

Input

5
2 9 8 2 7

Output

2 3
解题思路：简单贪心。女士从前往后贪心，男士从后往前贪心，即吃完当前位置的巧克力所花费时间较少者，便有优先往后（或往前）再吃一块巧克力，直到两者相遇即此时已经吃完了所有的巧克力，求女士和男士各吃了多少块巧克力。
AC代码：

#include<bits/stdc++.h>
using namespace std;
int n,first,last,sum=0,a,b,s[100005];
int main(){
    cin>>n;
    for(int i=1;i<=n;++i)cin>>s[i];
    a=s[(first=1)],b=s[(last=n)];
    while(first<=last){
        if(a<=b)a+=s[++first];//用时相等即两人刚好要吃相同的巧克力时，女士优先，男士靠边。
        else b+=s[--last];
    }
    cout<<(first-1)<<' '<<(n-last)<<endl;
    return 0;
}