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:
//No sorceries shall prevail.
import java.util.*;
import java.io.*;
public class Main {

	static Scanner inp= new Scanner(System.in);
	static PrintWriter out= new PrintWriter(System.out);
	public static void main(String args[]) {		    	
	   	Main g=new Main();
	   	g.main();
	   	out.close();
	}

	
	
	//Main
	void main() {
		
		//System.out.println(9*(long)Math.pow(10,18));
		int t=inp.nextInt();
		while(t-->0) {
			String s=inp.next();
			int x=inp.nextInt();
			int n=s.length();
			char a[]=s.toCharArray();
			long gg=0;
			int cur=n-1;
			while(sum(a,n)>x) {
				//System.out.println(a);
				gg+=make0(a,cur--,n);
			}
			out.println(gg);
		}
		
	}
	
	long make0(char s[], int index, int n) {
		if(s[index]=='0') return 0;
		long cost=(10-(s[index]-'0'))*(long)Math.pow(10, n-index-1);
		s[index]='0';
		index--;
		int rem=1;
		while(index>=0 && rem>0) {
			if(s[index]=='9') {
				s[index--]='0';
			}else {
				s[index]=(char)(s[index]+1);
				rem=0;
			}
		}
		//System.out.println(cost+"=cost");
		return cost;
	}
	
	int sum(char s[], int n) {
		int gg=0;
		for(int i=0;i<n;i++) {
			int x=s[i]-'0';
			gg+=x;
		}
		return gg;
	}
		
}