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:
//I AM THE CREED
/* //I AM THE CREED
/* package codechef; // don't place package name! */
import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.StringTokenizer; 
import java.util.*;
import java.awt.Point;
public class Main{
    //a.push_back
    
    public static void main(String[] args) throws IOException 
    { 
  
        Scanner input = new Scanner(System.in);
        input.nextInt();
        while(input.hasNext()){
            String s=input.next();
            long sum=input.nextLong();
            long ans=0;
            long[] n=new long[s.length()];
            long curr=0;
            for(int i=0;i<s.length();i++){
                n[i]=Integer.parseInt(s.substring(i, i+1));
                curr+=n[i];
            }
            long dig=0;
            for(int i=s.length()-1;i>=0;i--){
                if(n[i]==9 && dig==1){
                    n[i]=0;
                    dig=1;
                    curr-=9;
                    continue;
                }
                curr+=dig;
                n[i]+=dig;
                if(curr<=sum){
                    break;
                }
                ans+=(pow(s.length()-1-i)*(10-n[i]));
                curr-=n[i];
                dig=1;
                n[i]=0;
            }
            System.out.println(ans);
            
        }
    }
    public static long pow(int n){
        long ans=1;
        while(n!=0){
            ans=ans*10;
            n--;
        }
        return ans;
    }

}