Codeforces 706B 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 xicoins.

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

Copy

5
3 10 8 6 11
4
1
10
3
11

output

Copy

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.

这题目可以，题目说xi不超过100000，于是我就开了个数组用树状数组，提交runtime error，一看，哦，查询的是范围是1000000000，所以加了一步判断，这次是答案错误，看样子xi也是这么大的范围，这样一来，这题怕是通解另有其他，估计是二分，可是树状数组不行了吗，数组开不了特别大，那就map一下，过了，再写个二分，二分耗时少，果然是二分。。。

树状数组代码：

#include <iostream>
#include <map>
#include <cstdio>
#define MAX 1000000000
using namespace std;
int n,m;
map<int,int> sum;
int lowbit(int t) {
    return t&(-t);
}
void update(int x) {
    while(x <= MAX) {
        sum[x] ++;
        x += lowbit(x);
    }
}
int getans(int x) {
    int ans = 0;
    while(x > 0) {
        ans += sum[x];
        x -= lowbit(x);
    }
    return ans;
}
int main() {
    int d;
    scanf("%d",&n);
    for(int i = 0;i < n;i ++) {
        scanf("%d",&d);
        update(d);
    }
    scanf("%d",&m);
    for(int i = 0;i < m;i ++) {
        scanf("%d",&d);
        printf("%d
",getans(d));
    }
}

二分代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#define MAX 100000
using namespace std;
int n,m;
int x[MAX];

int main() {
    int d;
    scanf("%d",&n);
    for(int i = 0;i < n;i ++) {
        scanf("%d",&x[i]);
    }
    sort(x,x + n);
    scanf("%d",&m);
    for(int i = 0;i < m;i ++) {
        scanf("%d",&d);
        int l = 0,r = n,mid;
        while(l < r) {
            mid = (l + r) / 2;
            if(x[mid] <= d) l = mid + 1;
            else r = mid;
        }
        printf("%d
",l);
    }
}