Problem: You are given a positive integer n. In one move, you can increase n by one (i.e. make n := n + 1). Your task is to find the minimum number of moves you need to perform in order to make the sum of digits of n be less than or equal to s.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.

The only line of the test case contains two integers n and s (1 ≤ n ≤ 10^{18}; 1 ≤ s ≤ 162).

Output

For each test case, print the answer: the minimum number of moves you need to perform in order to make the sum of digits of n be less than or equal to s.

Example

Input


5
2 1
1 1
500 4
217871987498122 10
100000000000000001 1


Output


8
0
500
2128012501878
899999999999999999

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
import java.math.*;

   public class D
   {
      public static void main(String hi[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int T = Integer.parseInt(st.nextToken());
         StringBuilder sb = new StringBuilder();
         while(T-->0)
         {
            st = new StringTokenizer(infile.readLine());
            long N = Long.parseLong(st.nextToken());
            int S = Integer.parseInt(st.nextToken());
            if(N == 1000000000000000000L)
               sb.append("0\n");
            else
            {
               int[] digits = new int[19];
               for(int i=0; i < 19; i++)
               {
                  if(N == 0)
                     break;
                  digits[i] = (int)(N%10);
                  N /= 10;
               }
               long res = 0L;
               int sum = 0;
               for(int d: digits)
                  sum += d;
               if(sum > S)
               {
                  for(int d=0; d < 18; d++)
                  {
                     long temp = 10-digits[d];
                     for(int i=0; i < d; i++)
                        temp *= 10;
                     res += temp;
                     digits[d] = 0;
                     if(digits[d+1] == 9)
                     {
                        for(int i=d+1; i < 19; i++)
                        {
                           if(digits[i] == 9)
                              digits[i] = 0;
                           else
                           {
                              digits[i]++;
                              break;
                           }
                        }
                     }
                     else
                        digits[d+1]++;
                     sum = 0;
                     for(int dd: digits)
                        sum += dd;
                     if(sum <= S)
                        break;
                  }
               }
               sb.append(res+"\n");
            }
         }
         System.out.print(sb);
      }
   }