Algorithm


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
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 123 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.

 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include<bits/stdc++.h>
using namespace std;


#define ll long long
#define endl '\n'
#define debug(n) cout<<(n)<<endl;
const ll INF = 2e18 + 99;


int main(){
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);

  int n;
  cin>>n;
  int arr[n];
  for(int i = 0; i < n; i++){
    cin>>arr[i];
  }
  int m;
  cin>>m;
  sort(arr, arr+n);
  while(m--){
    int t;
    cin>>t;
    int a;
    a = upper_bound(arr, arr+n, t) - arr;
    cout<<a<<endl;


  }

}
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Input

x
+
cmd
5
3 10 8 6 11
4
1
10
3
11

Output

x
+
cmd
0
4
1
5
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Demonstration


Codeforcess Solution 706-B B. Interesting drink ,C++, Java, Js and Python ,706-B,Codeforcess Solution

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