codeforces#271 (Div. 2)预处理

B. Worms

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

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.
• 预处理解决,防止超时

• #include<bits/stdc++.h>
  using namespace std;
  int a[1000005];
  int main()
  {
  	int n,n1;
  	int c,b;
  	int cnt=0;
  	cin>>n;
  	for(int i=1;i<=n;i++)
  	{
  		cin>>b;
  		for(int j=1;j<=b;j++)
  		{
  			a[++cnt]=i;
  		}
  	}
  	cin>>n1;
  	for(int i=1;i<=n1;i++)
  	{
  		cin>>c;
  		cout<<a[c]<<endl;
  	}
  }