Vasiliy plans to buy his favorite drink for q consecutive days. He knows, that on the i-th day he will be able to spent mi coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".

</big>##Input
<big>

The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of shops in the city that sell Vasiliy’s favourite drink.

The second line contains n integers xi (1 ≤ xi ≤ 100 000) — prices of the bottles of the drink in the i-th shop.

The third line contains a single integer q (1 ≤ q ≤ 100 000) — the number of days Vasiliy plans to buy the drink.

Then follow q lines each containing one integer mi (1 ≤ mi ≤ 109) — the number of coins Vasiliy can spent on the i-th day.

</big>##Output
<big>

Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.

</big>##Examples
<big>
input
5
3 10 8 6 11
4
1
10
3
11
output
0
4
1
5
</big>##Source
<big>
Codeforces Round #367 (Div. 2)
</big><br/>
##题意：
<big>
给出一种饮料在不同店的价格，给出每天的消费额度，求每天可以选择多少个店.
</big><br/>
##题解：
<big>
预处理一个前缀和即可.
</big><br/>
##代码：
``` cpp
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <vector>
#include <list>
#define LL long long
#define eps 1e-8
#define maxn 101000
#define mod 100000007
#define inf 0x3f3f3f3f
#define mid(a,b) ((a+b)>>1)
#define IN freopen("in.txt","r",stdin);
using namespace std;int cnt[maxn];int main(int argc, char const *argv[])
{
    //IN;    int n;
    while(scanf("%d", &n) != EOF)
    {
        memset(cnt, 0, sizeof(cnt));
        for(int i=1; i<=n; i++) {
            int x; cin >> x;
            cnt[x]++;
        }        for(int i=1; i<maxn; i++)
            cnt[i] += cnt[i-1];        int q; cin >> q;
        while(q--) {
            int x; cin >> x;
            if(x >= maxn-1) printf("%d\n", cnt[maxn-1]);
            else printf("%d\n", cnt[x]);
        }
    }    return 0;
}