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.Scanner;

public class DecreaseSumDigits {

	public static void main(String[] args) {
		
		Scanner sc = new Scanner(System.in);
		
		int tests = sc.nextInt();
		
		for(int t=0; t<tests; t++) {
			long n = sc.nextLong();
			long s = sc.nextLong();
			
			long sum = 0;
			long ans = 0;
			long curr = 0;
			
			long divisor = (long)Math.pow(10,18);
			
			if(divisor==n || (n<10 && n<=s)) {
				System.out.println(0);
				continue;
			}
			if(n<10 && n>s) {
				System.out.println(10-n);
				continue;
			}
			while(n/divisor==0) {
				divisor/=10;
			}
			divisor*=10;
			curr=1*divisor;
			ans=curr-n;
			divisor/=10;
			curr=0;
			long num = n;
			long digit = 0;
			while(num>0) {
				digit = num/divisor;
				digit++;
				sum+=digit;
				if(sum>s) {
					break;
				}
				else {
					curr+=(digit*divisor);
					ans=Math.min(ans,curr-n);
					curr-=(digit*divisor);
					digit--;
					curr+=(digit*divisor);
					sum--;
					num=num%divisor;
					//System.out.println(num);
					divisor=divisor/10;
				}
			}
			if(num==0) {
				System.out.println(0);
			}
			else {
				sum--;
				if(sum==s) {
					curr+=(--digit*divisor);
					if(curr-n==0) {
						ans=0;
					}
				}
				System.out.println(ans);
			}
			
		}
		sc.close();

	}

}