二分

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 x_i 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 m_i 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 x_i (1 ≤ x_i ≤ 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 m_i (1 ≤ m_i ≤ 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.

二分水题

给出一个序列的长度和元素  然后有m个数  分别求出这个序列中小于等于这些数的元素个数  

#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
int a[100001];
int main()
{
     int n;
     scanf("%d",&n);
     for(int i=0;i<n;i++)
     {
     	scanf("%d",&a[i]);
	 }
	 sort(a,a+n);
	 int m,k;
	 scanf("%d",&m);
	 while(m--)
	 {
       scanf("%d",&k);
       int r=n-1,l=0;
       int ans=0;
       while(l<=r)
       {
       	int mid=(l+r)/2;
		   if(a[mid]>k)
		   {
		   	r=mid-1;
		   }
		   else
		   {
		   	  l=mid+1  ;
		   }
	   }
	   printf("%d\n",l);
	 }
	return 0;
}