题目

Description

Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in n different shops in the city. It's known that the price of one bottle in the shop i is equal to xi coins.

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".

Input

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.

Output

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.

Sample Input

Input

5
3 10 8 6 11
4
1
10
3
11

Output

0
4
1
5

Hint

On the first day, Vasiliy won't be able to buy a drink in any of the shops.

On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.

On the third day, Vasiliy can buy a drink only in the shop number 1.

Finally, on the last day Vasiliy can buy a drink in any shop.

题解

多次查询小于等于 m 的数据的个数
查询次数较多,直接查找会超时
应该使用二分查找
使用 upper_bound() 可以直接获得答案

upper_bound 和 lower_bound

upper_bound(a,a+n,e)

返回 a ~ a+n 区域内大于 e 的第一个内存地址
用其减去首地址即可得到下标
则可以直接获取大于 e 的第一个数据的下标
如果从 0 开始的话,恰好多的一位补到从 0 开始错的一位上

lower_bound(a,a+n,e)

返回 a ~ a+n 区域大于等于 e 的第一个内存地址
如果 e 不存在,则 lower_bound() 和 upper_bound() 的结果应该是一样的
如果有 e 存在,则 lower_bound() 会返回第一个 e 的地址
而 upper_bound() 会返回最后一个 e 的下一个地址

代码

/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <functional>
#include <bitset>
#include <iomanip> 
using namespace std;

const int maxn = 100005;
int a[maxn];

bool Do() {
    int n;
    if(!(cin >> n))
        return false;
    for(int i = 0;i < n;i++)
        cin >> a[i];

    sort(a,a + n);

    int q;
    cin >> q;
    for(int i = 0;i < q;i++) {
        int t;
        cin >> t;
        cout << upper_bound(a,a + n,t) - a << endl;
    }

    return true;
}

int main() {
    cin.tie(0);
    cin.sync_with_stdio(false);

    while(Do());
    return 0;
}