Codeforces Round #271 (Div. 2)

It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.

Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.

Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.

Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.

Input

The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.

The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.

The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.

Output

Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.

Sample test(s)

input

5
2 7 3 4 9
3
1 25 11

output

1
5
3

题意：给你n个序列的长度，依次连接n个序列，从第一个開始计数。

求第m个原本属于那个序列

思路：记录每一个数属于那个小序列即可了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1e6;

int num[maxn];

int main() {
	int n;
	scanf("%d", &n);
	int cnt = 0;
	int m;
	for (int i = 1; i <= n; i++) {
		scanf("%d", &m);
		for (int j = 0; j < m; j++)
			num[++cnt] = i;
	}
	int q;
	scanf("%d", &q);
	while (q--) {
		scanf("%d", &m);
		printf("%d
", num[m]);
	}
	return 0;
}