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 Task{
// taking inputs
static BufferedReader s1;
static BufferedWriter out;
static String read() throws IOException{String line="";while(line.length()==0){line=s1.readLine();continue;}return line;}
static int int_v (String s1){return Integer.parseInt(s1);}
static long long_v(String s1){return Long.parseLong(s1);}
static int[] int_arr() throws IOException{String[] a=read().split(" ");int[] b=new int[a.length];for(int i=0;i<a.length;i++){b[i]=int_v(a[i]);}return b;}
static long[] long_arr() throws IOException{String[] a=read().split(" ");long[] b=new long[a.length];for(int i=0;i<a.length;i++){b[i]=long_v(a[i]);}return b;}
static void assign(){s1=new BufferedReader(new InputStreamReader(System.in));out=new BufferedWriter(new OutputStreamWriter(System.out));}
static long gcd(long a,long b){if(b==0){return a;}return gcd(b,a%b);}
static long Modpow(long a,long p,long m){long res=1;while(p>0){if((p&1)!=0){res=(res*a)%m;}p >>=1;a=(a*a)%m;}return res;}
static long Modmul(long a,long b,long m){return ((a%m)*(b%m))%m;}
static long ModInv(long a,long m){return Modpow(a,m-2,m);}

//sort map w.r.t value use-> TreeSet of array? write comparator which depends on all the entries of array otherwise weired behavior you can prove it.
//......................................@uthor_Alx..............................................
   static int f(long n){
        int s=0;
        while(n>0){
            s+=(int)(n%10l);n/=10;
        }
        return s;
   }
   public static void main(String[] args) throws  IOException{
                assign();
                int t=int_v(read()),cn=1;
                while(t--!=0){
                    long[] x1=long_arr();
                    long n=x1[0],s=x1[1];
                    long sum=0;long c=1l;
                    while(f(n)>s){
                        int r=(int)(n%10l);
                        long kk=(10l-r);kk%=10l;kk*=c;
                        sum+=kk;
                        n/=10;c*=10;
                        if(r!=0) n++;
                    }
                    out.write(sum+"\n");
                }
                    out.flush();

    }
}