Cinema in Akiba (CIA) is a small but very popular cinema in Akihabara. Every night the cinema is full of people. The layout of CIA is very interesting, as there is only one row so that every audience can enjoy the wonderful movies without any annoyance by other audiences sitting in front of him/her.

The ticket for CIA is strange, too. There are n seats in CIA and they are numbered from 1 to n in order. Apparently, n tickets will be sold everyday. When buying a ticket, if there are k tickets left, your ticket number will be an integer i (1 ≤ i ≤ k), and you should choose the ith empty seat (not occupied by others) and sit down for the film.

On November, 11th, n geeks go to CIA to celebrate their anual festival. The ticket number of the ith geek is a_i. Can you help them find out their seat numbers?

Input

The input contains multiple test cases. Process to end of file.

The first line of each case is an integer n (1 ≤ n ≤ 50000), the number of geeks as well as the number of seats in CIA. Then follows a line containing n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ n - i + 1), as described above. The third line is an integer m (1 ≤ m ≤ 3000), the number of queries, and the next line is m integers, q₁, q₂, ..., q_m (1 ≤ q_i ≤ n), each represents the geek's number and you should help him find his seat.

Output

For each test case, print m integers in a line, seperated by one space. The ith integer is the seat number of the q_ith geek.

Sample Input

31 1 131 2 352 3 3 2 152 3 4 5 1

Sample Output

1 2 34 5 3 1 2

题意：就是一排人看电影，票上安排给某人的位置是第几个空位。让你找出某些人的相应座位号。

线段树保存空位个数进行处理。

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          typedef long long LL;using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )const int maxn=50000+100;int sum[maxn<<2];int ans[maxn];int n,m;void pushup(int rs){    sum[rs]=sum[rs<<1]+sum[rs<<1|1];}void build(int rs,int l,int r){    sum[rs]=r-l+1;    if(l==r)  return ;    int mid=(l+r)>>1;    build(rs<<1,l,mid);    build(rs<<1|1,mid+1,r);    pushup(rs);}int update(int rs,int l,int r,int x){ //    cout<<"2333 "<
          
           <<" "<
           
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             >1; if(x<=sum[rs<<1]) ret=update(rs<<1,l,mid,x); else ret=update(rs<<1|1,mid+1,r,x-sum[rs<<1]); pushup(rs); return ret;}int main(){ int x; while(~scanf("%d",&n)) { CLEAR(sum,0); CLEAR(ans,0); build(1,1,n); REPF(i,1,n) { scanf("%d",&x); ans[i]=update(1,1,n,x); } scanf("%d",&m); while(m--) { scanf("%d",&x); printf(m==0?"%d\n":"%d ",ans[x]); } } return 0;}