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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Bisoye
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        DDecreaseTheSumOfDigits solver = new DDecreaseTheSumOfDigits();
        int testCount = Integer.parseInt(in.next());
        for (int i = 1; i <= testCount; i++)
            solver.solve(i, in, out);
        out.close();
    }

    static class DDecreaseTheSumOfDigits {
        public void solve(int testNumber, Scanner in, PrintWriter out) {

            long n = in.nextLong(), s = in.nextInt();
            if (sumOfDigit(n) <= s) {
                out.println(0);
                return;
            }

            long pw = 1;
            long old = n;
            while (sumOfDigit(n) > s) {
                pw *= 10;
                n /= 10;
                n++;
            }

            out.println(n * pw - old);


        }

        public int sumOfDigit(long digit) {
            int res = 0;
            while (digit > 0) {
                res += digit % 10;
                digit /= 10;
            }
            return res;
        }

    }
}