CodeForces

Interesting drink

Problem

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.

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.

给出n个数，再输入若干个M，每输入一次M，求n个数里有多少个数小于M。
由于n是1e5的数量级，不可O(n^2)暴力，用二分即可。

#include<stdio.h>  
    #include<string.h>  
    #include<algorithm>  
    using namespace std;  
    int a[100000+10];  
    int main()  
    {  
        int n,i,j,k,num,m;  
        while(scanf("%d",&n)!=EOF)  
        {  
            for(i=1;i<=n;i++)  
            scanf("%d",&a[i]);  
            sort(a+1,a+n+1);  
            scanf("%d",&m);  
            while(m--)  
            {  
                scanf("%d",&num);  
                if(num<a[1])  
                printf("0
");  
                else if(num>=a[n])  
                printf("%d
",n);  
                else  
                {  
                    int left=1;  
                    int right=n;  
                    int mid;  
                    int ans;  
                    while(left<=right)  
                    {  
                        mid=(left+right)/2;  
                        if(a[mid]<=num)  
                        {  
                            ans=mid;  
                            left=mid+1;  
                        }  
                        else  
                        {  
                            right=mid-1;  
                        }  
                    }  
                    printf("%d
",ans);  
                }  
            }  
        }  
        return 0;  
    }