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.*; 
  
public class D 
{ 
    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; 
        } 
    } 
    static final FastReader in = new FastReader();
    long tc(){
        long n = in.nextLong(), s = in.nextInt();
        String str = n+"";
        long pres = 0;
        for(int i=0; i<str.length(); i++){
            int d = str.charAt(i)-'0';
            pres+=d;
            if(pres>s){
                pres++;
                int j = i;
                for(; j>=0; j--){
                    d = str.charAt(j)-'0';
                    pres-=d;
                    if(pres<=s)
                    break;
                }
                if(j==0)
                return (long)Math.pow(10l, str.length())-n;
                else
                {
                    return (Long.parseLong(str.substring(0,j))+1l)*(long)Math.pow(10l, str.length()-j)-n;
                }
            }
        }
        return 0;
    }
    public static void main(String[] args) 
    { 
        StringBuilder out = new StringBuilder();
        D ob = new D();
        int t = in.nextInt();
        while(t-->0){
           out.append(ob.tc()+"\n");
        }
        System.out.print(out);
    } 
}