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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class dsaaaaaaaaaaaaaaaaaaaaaaaaaaaa {
	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; 
        } 
    } 
	public static void main(String[] args) {
		FastReader s=new FastReader();
         int t=s.nextInt();
         while(t>0) {
        	 long n=s.nextLong();
        	 int v=s.nextInt();
        	String g=String.valueOf(n);
        	int m=0;
        	for(int i=0;i<g.length();i++) {
        		m+=g.charAt(i)-'0';
        	}
        	int h=0;
        	int i=0;
        	boolean p=false;
        	for( i=0;i<g.length();i++) {
        		h+=g.charAt(i)-'0';
        		if(h>v) {
        			p=true;
        			break;
        		}else if(h==v &&i!=g.length()-1 && m>v) {
        			p=true;
        			break;
        		}
        	}
        	if(!p) {
        		System.out.println(0);
        	}
        	else {
        	
        		int j=0;
        		String k="";
        		boolean u=false;
        		for(j=g.length()-1;j>=i;j--) {
        		if(j==g.length()-1 && g.charAt(j)!='0') {
        			u=true;
        			k+=10-(g.charAt(j)-'0');
        		}else if(j==g.length()-1 && g.charAt(j)=='0') {
        			k+='0';
        		}else {
        			if(g.charAt(j)!='0' && !u) {
        				u=true;
        				k+=10-(g.charAt(j)-'0');
        			}else if(g.charAt(j)!='0' && u) {
        				k+=9-(g.charAt(j)-'0');
        			}else if(g.charAt(j)=='0' && !u) {
        				k+='0';
        			}else {
        				k+='9';
        			}
        		}
        			
        		}
        		boolean v1=false;
        		        		for(int b=k.length()-1;b>=0;b--) {
        			if(k.charAt(b)!='0') {
        				v1=true;
        			}
        			if(k.charAt(b)=='0' && !v1) {
        				continue;
        			}
        		      System.out.print(k.charAt(b));
        		}
        		System.out.println();
        	}
         t--;
         }

	}
	}