B. Interesting drink

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本文介绍了一道关于购买特定饮料的问题，通过排序与二分查找的方法高效地解决了主人公连续多天购买饮料的问题。该算法首先对商店的价格进行排序，然后针对每天的预算使用二分查找确定可购买饮料的商店数量。

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.

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

先排序，后二分！！！

#include <iostream>
#include <bits/stdc++.h>

using namespace std;

int a[100005];
int main()
{
    int n,x,pos,q;
    long long mi;
    cin>>n;
    for(int i=0; i<n; i++)
    {
        scanf("%d",&a[i]);
    }
    cin>>q;
    sort(a,a+n);
    while(q--)
    {
        scanf("%d",&x);
        if(a[0]>x)
            cout<<'0'<<endl;
        else if(a[n-1]<x)
            cout<<n<<endl;
        else
        {
            int head=0,tail=n-1,mid;
            while(head<=tail)
            {
                mid=(head+tail)/2;
                if(a[mid]<=x)
                {
                    pos=mid;
                    head=mid+1;
                }
                else
                {
                    tail=mid-1;
                }
            }
            cout<<pos+1<<endl;
        }
    }
    return 0;
}