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.*;
import java.io.*;
public class D {
	public static void main(String args[]) {
		FastScanner in = new FastScanner();
		int t=in.nextInt();
		L :while(--t>=0) {
			long num=in.nextLong();
			int sum=in.nextInt();
			
			char s[]=Long.toString(num).toCharArray();
			int n=s.length;
			long copy_num=num;
			
			if(countSum(num)<=sum) {
				System.out.println(0);
				continue L;
			}
			for(int i=n-1;i>=0;i--) {
				if(s[i]!='0') {
					long add=(long)Math.pow(10, n-1-i)*(10-(s[i]-'0'));
					num=num+add;
					s=Long.toString(num).toCharArray();
				}
				if(countSum(num)<=sum) {
					System.out.println(num-copy_num);
					continue L;
				}
			}
		}
	}
	
	static int countSum(long num) {
		int sum=0;
		while(num>0) {
			sum+=(int)(num%10);
			num=num/10;
		}
		return sum;
	}

	///////////////////////////
	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");

		String next() {
			while (!st.hasMoreTokens())
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		int[] nextArray(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}

		long nextLong() {
			return Long.parseLong(next());
		}
	}
}