Polycarp is reading a book consisting of n pages numbered from 1 to n. Every time he finishes the page with the number divisible by m, he writes down the last digit of this page number. For example, if n=15 and m=5, pages divisible by m are 5, 10, 15. Their last digits are 5, 0, 5 correspondingly, their sum is 10.

Your task is to calculate the sum of all digits Polycarp has written down.

You have to answer q independent queries.

Input

The first line of the input contains one integer q (1 ≤ q ≤ 1000) — the number of queries.

The following q lines contain queries, one per line. Each query is given as two integers n and m (1 ≤ n, m ≤ 10^{16}) — the number of pages in the book and required divisor, respectively.

Output

For each query print the answer for it — the sum of digits written down by Polycarp.

Example

Input


7
1 1
10 1
100 3
1024 14
998244353 1337
123 144
1234312817382646 13


Output


1
45
153
294
3359835
0
427262129093995

import java.io.*;
import java.util.*;
import java.lang.*;
import java.math.BigInteger;
public class abc {
    
    public static void main(String args[])throws IOException
    {
        BufferedReader br=new BufferedReader(new InputStreamReader (System.in));
        
        int t=Integer.parseInt(br.readLine());
        int i=0;
        while(t-->0)
        {
        StringTokenizer tk=new StringTokenizer(br.readLine());
        long n=Long.parseLong(tk.nextToken());
        long m=Long.parseLong(tk.nextToken());
        long p=0,sum=0;
        if(m%10==0)
        System.out.println("0");
        else if(m%5==0)
        {
            long a[]=new long[2];
            a[0]=5;
            a[1]=0;
            sum=5;
            p=n/m;
            if(p%2==0)
            {
                sum=sum*(p/2);
                System.out.println(sum);
            }
            else 
            {
                sum=((p-1)/2)*sum+5;
                System.out.println(sum);
            }
        }
        else if(m%2==0)
        {
            long b[]=new long[5];sum=0;
            for(i=0;i<5;i++)
            {
                p=m*(long)(i+1);
                b[i]=p%10;
                sum=sum+(long)b[i];
            }
            p=n/m;
            
                sum=(p/5)*sum;
                for(i=0;i<(p%5);i++)
                sum=sum+b[i];
                System.out.println(sum);
            
        }
        else
        {
              long c[]=new long[10];sum=0;
            for(i=0;i<10;i++)
            {
                p=m*(long)(i+1);
                c[i]=p%10;
                sum=sum+(long)c[i];
            }
            p=n/m;
            
                sum=(p/10)*sum;
                for(i=0;i<(p%10);i++)
                sum=sum+c[i];
                System.out.println(sum);
            
        }
      
    }
}
}