浙南联合训练赛20180318

英语水平大考验

B - Worms

CodeForces - 474B

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

Marmot brought Molenordered piles of worms such thati-th pile containsa_iworms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers1toa₁, worms in second pile are labeled with numbersa₁ + 1toa₁ + a₂and 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 integern(1 ≤ n ≤ 10⁵), the number of piles.

The second line containsnintegersa₁, a₂, ..., a_n(1 ≤ a_i ≤ 10³,a₁ + a₂ + ... + a_n ≤ 10⁶), wherea_iis the number of worms in thei-th pile.

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

The fourth line containsmintegersq₁, q₂, ..., q_m(1 ≤ q_i ≤ a₁ + a₂ + ... + a_n), the labels of the juicy worms.

Output

Printmlines to the standard output. Thei-th line should contain an integer, representing the number of the pile where the worm labeled with the numberq_iis.

Example

Input

5<br />2 7 3 4 9<br />3<br />1 25 11

Output

1<br />5<br />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.

就是给你的集合的前面让你二分的

#include <bits/stdc++.h>
using namespace std;
const int N=1e5+;
int a[N];
int main()
{
    int n,q;
    cin>>n;
    a[]=;
    for(int i=,x;i<=n;i++)
        cin>>x,a[i]=a[i-]+x;
    cin>>q;
    for(int i=,x;i<=q;i++)
    {
        cin>>x;
        int pos=lower_bound(a,a+n+,x)-a;
        if(a[pos]==x)pos++;
        cout<<pos<<"\n";
    }
}