Day 30 B. Worms

Problem
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 ≤ 10^5), the number of piles.

The second line contains n integers a1, a2, …, an (1 ≤ ai ≤ 10^3, a1 + a2 + … + an ≤ 10^6), where ai is the number of worms in the i-th pile.

The third line contains single integer m (1 ≤ m ≤ 10^5), 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.

Examples

input
5
2 7 3 4 9
3
1 25 11

output
1
5
3

Note

For the sample input:
The worms with labels from [1, 2] are in the first pile.
The worms with labels from [3, 9] are in the second pile.
The worms with labels from [10, 12] are in the third pile.
The worms with labels from [13, 16] are in the fourth pile.
The worms with labels from [17, 25] are in the fifth pile.

题目大致意思为：给n个数字 分成n段 每段区间范围为an 再给出m次询问 问qn在哪个区间分段里 输出第几分段

#include
#include
#include
#include
#include
#include
#include
#include
#include
#define ll unsigned long long
using namespace std;
int m[2000005];
//开双重循环记录每段区间内容 再查询 暴力写法 不是最优化的
int main()
{
	int n;
	int cnt = 0;
	cin >> n;
	for (int i = 0; i < n; i++)
	{
		int x;
		cin >> x;
		for (int j = 0; j < x; j++)
		{
			m[++cnt] = i + 1;
		}
	}
	int t;
	cin >> t;
	while (t--)
	{
		int x;
		cin >> x;
		cout << m[x] << endl;
	}
	return 0;
}