codeforces 706B B. Interesting drink (二分)

B. Interesting drink

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

53 10 8 6 114110311

output

0415

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.

题目大意：只要那天的价格小于主人的钱数就行，求多少个符合的商店

思路：先sort排序,数据大，逐个查找o(n*q),用二分查找商店，下标就是答案

代码：

#include<cstdio>#include<algorithm>using namespace std;int a[100000+10]; int main(){int n,m,x;while(~scanf("%d",&n)){for(int i=1;i<=n;i++){scanf("%d",&a[i]);}sort(a+1,a+n+1); scanf("%d",&m);int num;for(int i=0;i<m;i++){scanf("%d",&x);num=0;int pos=upper_bound(a+1,a+n+1,x)-a;printf("%d\n",pos-1);}}return 0;}