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.io.*;
import java.util.*;
import java.awt.*;
public class A
{
    BufferedReader in;
    PrintWriter ob;
    StringTokenizer st;
 
    public static void main(String[] args) throws IOException {
        new A().run();
    }
    void run() throws IOException {
        in=new BufferedReader(new InputStreamReader(System.in));
        ob=new PrintWriter(System.out);
        int t = ni();
        while(t-->0)
        solve();
        ob.flush();
    }
    void solve() throws IOException {
        long a = nl() , b = nl();
        if( a%b == 0 )
            ob.println("0");
        else {
            long extra = ((a/b) + 1)*b;
            ob.println(extra - a);
        }
    }
 
    String ns() throws IOException {
        return nextToken();
    }
    long nl() throws IOException {
        return Long.parseLong(nextToken());
    }
    int ni() throws IOException {
        return Integer.parseInt(nextToken());
    }
    double nd() throws IOException {
        return Double.parseDouble(nextToken());
    } 
    String nextToken() throws IOException {
        if(st==null || !st.hasMoreTokens())
            st=new StringTokenizer(in.readLine());
        return st.nextToken();
    }
    int[] nia(int start,int b) throws IOException {
        int a[]=new int[b];
        for(int i=start;i<b;i++)
            a[i]=ni();
        return a;
    }
    long[] nla(int start,int n) throws IOException {
        long a[]=new long[n];
        for (int i=start; i<n ;i++ ) {
            a[i]=nl();
        }
        return a;
    }
}