codeforces706B Interesting drink 排序+二分查找

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本文介绍了一个程序员为了在连续的几天内用有限的资金购买心爱的饮料“Beecola”，需要解决的问题是如何确定每天可以在多少家商店买到这种饮料。通过排序和二分查找的方法，文章提供了一种有效的解决方案。

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.

Example

input
5
3 10 8 6 11
4
1
10
3
11

output
0
4
1
5

Note

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.

从大到小排个序，二分查找比输入的数大的第一个数，然后这个数的位置下标减一即输出

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <cstdlib>
#include <algorithm>

using namespace std;

const int maxn = 100000;
int a[maxn + 5];

int main()
{
	int n, q, tmp;
	cin >> n;
	for (int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
	}
	sort(a + 1, a + n + 1);
	cin >> q;
	while (q--) {
		scanf("%d", &tmp);
		int l, r;
		l = 1; r = n;
		while (l <= r) {
			int mid = (l + r) / 2;
			if (a[mid] > tmp) {
				r = mid - 1;
			}
			else {
				l = mid + 1;
			}
		}
		printf("%d\n", l - 1);
	}
	return 0;
}