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, …, tn (1 ≤ ti ≤ 1000), where ti 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
#include<stdio.h>
int main()
{
int n,a[100008],l,m;
int sum=0,sum1=0;
scanf("%d",&n);
l=0;m=n-1;//l和m分别是Alice和Bob的吃的巧克力棒的下标数
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
while(l<=m)//循环的条件，之所以有'=',是两人同时应该吃到一块巧克力棒
{
if(sum<=sum1)//‘=’当俩人同时应该吃同一个巧克力时，这块巧克力最后是被Alice吃
sum+=a[l++];
else
{
sum1+=a[m--];
}
}
printf("%d %d",l,n-l);
return 0;
}