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

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.

 

这一场本来没什么要讲的，直到刚刚闲着无聊翻了翻巨巨们的代码！！

 

题意：
找出对于给定数m在x中有多少数不大于m、

 

简单的二分查找也可以写的漂亮！！

现学现卖的用了一下upper_bound()函数，简直方便，只不过比直接二分要慢。

 

关于upper_bound()函数请见我的下一篇帖子，有详细的介绍：传送门

 

附AC代码：

//  685 ms    2400 KB
#include<bits/stdc++.h>
using namespace std;

int a[100010];

int main(){
    int n,q,x;
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    sort(a,a+n);
    cin>>q;
    while(q--){
        cin>>x;
        int res=upper_bound(a,a+n,x)-a;//upper_bound()返回的是迭代器的位置 
        cout<<res<<endl;
    }
    return 0;
} 

 

 

//  78 ms    2200 KB 
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int a[100010];
int main()
{
    int n,m;
    while(scanf("%d",&n)!=EOF)
    {
        int b,i,j;
        for(i = 1 ; i <= n ; i++)
        {
            scanf("%d",&a[i]);
        }
        sort(a+1,a+n+1);
        scanf("%d",&m);
        for(i = 1 ; i <= m ; i++)
        {
            scanf("%d",&b);
            int l=1,r=n,mid,num=0;
            while(l <= r)
            {
                mid=(l+r)/2;
                if(a[mid] > b)
                    r=mid-1;
                else
                {
                    num=mid;
                    l=mid+1;
                }        
//                printf("%d------ %d\n",l,r);
            }
            printf("%d\n",num);
        }
    }
}