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.util.*;
 
public class D 
{	
	public static void main(String[] args)throws IOException 
	{	
		FastReader f=new FastReader();
		StringBuffer sb = new StringBuffer();
		
		int test=f.nextInt();
		out:
		while(test-->0)
		{	
			long n=f.nextLong();
			int s=f.nextInt();
			
			int no=0;
			long cpy=n;
			while(cpy>0)
			{	
				cpy/=10;
				no++;
			}
			if(sum(n)<=s)
			{
				sb.append("0\n");
				continue out;
			}
			long ans=0;
			long pow=1;
			for(int i=0;i<no;i++)
			{
				int digit= (int)((n/pow)%10);
				long add=pow*((10-digit)%10);
				n+=add;
				ans+=add;
				if(sum(n)<=s)
					break;
				pow*=10;
			}
			sb.append(ans+"\n");
		}
		System.out.println(sb);
	}
	
	static int sum(long n)
	{	
		int sum=0;
		while(n>0)
		{
			sum+=n%10;
			n/=10;
		}
		return sum;
	}
	
	static class FastReader 
	{ 
	    BufferedReader br; 
	    StringTokenizer st; 

	    public FastReader() {
	    	br = new BufferedReader(new
	                 InputStreamReader(System.in)); 
	    }
	    String next() { 
	        while (st == null || !st.hasMoreElements()) { 
	            try{ 
	                st = new StringTokenizer(br.readLine()); 
	            } 
	            catch (IOException  e) { 
	                e.printStackTrace(); 
	            } 
	        } 
	        return st.nextToken(); 
	    } 
	    int nextInt() { 
	        return Integer.parseInt(next()); 
	    } 
	    long nextLong() { 
	        return Long.parseLong(next()); 
	    } 
	    double nextDouble() { 
	        return Double.parseDouble(next()); 
	    } 
	    String nextLine() { 
	        String str = ""; 
	        try{ 
	            str = br.readLine(); 
	        } 
	        catch (IOException e) { 
	            e.printStackTrace(); 
	        } 
	        return str; 
	    } 
	} 
}