Worms(水题）

Description

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 thati-th pile contains a_i worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers1 to a₁, worms in second pile are labeled with numbersa₁ + 1 toa₁ + 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, ..., a_n (1 ≤ a_i ≤ 10³,a₁ + a₂ + ... + a_n ≤ 10⁶), where a_i is the number of worms in thei-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, ..., q_m (1 ≤ q_i ≤ a₁ + a₂ + ... + a_n), the labels of the juicy worms.

Output

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

Sample Input

Input

52 7 3 4 931 25 11

Output

153

Sample Output

Hint

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.

题意：第一行n为输入的区间数，第二行为N个区间的每个区间的长度，第三行m词询问，询问m个数在第几个区间里

ps：坑死了，TLE了四次，只因为数组没拿出来，到现在也没弄明白才10^6为嘛就要拿出来，飞神说超过1000的数组他就放在外面，以后就这么干了。

#include <stdio.h>#include <string.h>#include <stdlib.h>int num[1000010];int main(){    int T,n,m,p;    int i,j;    int cnt=1;    while(~scanf("%d",&T))    {        for(i=1; i<=T; i++)        {            scanf("%d",&n);            for(j=1; j<=n; j++)                num[cnt++]=i;        }        scanf("%d",&m);        for(i=1; i<=m; i++)        {            scanf("%d",&p);            printf("%d\n",num[p]);        }    }    return 0;}