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.lang.reflect.Array;
import java.util.*;

public class Main {
    public static long sum(long n) {
        long res = 0;
        while (n > 0) {
            res += n % 10;
            n /= 10;
        }
        return res;
    }


    public static void main(String[] args) throws IOException {
//        BufferedReader br = new BufferedReader(
//                new InputStreamReader(System.in));
//        BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
//        log.flush();
        Scanner sc = new Scanner(System.in);
        int test = sc.nextInt();
        for (int t = 0; t < test; t++) {
            long n = sc.nextLong(), s = sc.nextLong();
            if (sum(n) <= s) {
                System.out.println(0);
                continue;
            }
            long ans = 0;
            long pw = 1;
            for (int i = 0; i < 18; i++) {
                long digit = (n / pw) % 10;
                long add = ((10 - digit) % 10) * pw;
                n += add;
                ans += add;
                if (sum(n) <= s)
                    break;
                pw *= 10;
            }
            System.out.println(ans);
        }
    }
}
//x1 + x2 + ... + xn = s;

// a2−a1=a3−a2=a4-a3
//1234
//2-1=3-2
//3-2=4-3
// 2^n*n + 2^(n-1)*(n-1) + ... + 2^3 +