#include <iostream>

using namespace std;
int main()
 {
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int m,n;
      scanf("%d",&n);
      int count=0;
      for(int i=5; ;i=i+5)
      {
          m=i;
          while(m%5==0 && m>0)
          {
              m=m/5;
              count++;
              if(count==n)
              {
                  break;
              }
          }
          if(count==n)
          {
              printf("%d",i);
              break;
          }
      }
      printf("\n");
    }
     

	return 0;
}

Problem Description

Given a number n. The task is to find the smallest number whose factorial contains at least n trailing zeroes.

Examples:

Input : n = 1
Output : 5
1!, 2!, 3!, 4! does not contain trailing zero.
5! = 120, which contains one trailing zero.

Input : n = 6
Output : 25
25! = 15511210043330985984000000

Input:
The first line of input contains an integer T denoting the no of test cases. Then T test cases follow. Each test case contains an integer n.

Output:
For each test case in a new line print the required output.

Constraints:
1<=T<=100
1<=n<=100

• Test Case 1

  Input (stdin)

  2
  1
  6
  

  Expected Output

  5
  25

• Test Case 2

  Input (stdin)

  3
  2
  5
  

  Expected Output

  10
  25
  25