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:
                ////////////////////////////////////////////////////
                //                                                //
                //  For her who keeps the fire kindling in me...  //
                //                                                //
                ////////////////////////////////////////////////////


import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.FileOutputStream;
import java.io.IOException; 
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
// import java.util.Scanner;
// import java.util.Arrays;
// import java.util.ArrayList;
// import java.util.Collections; 
// import java.util.StringTokenizer;
import java.util.*;
import java.lang.Math;
import java.sql.Array;
import java.awt.Point;                            

import java.awt.geom.*;

public class Main{

    static class Node{
        int x;
        int rank;
        int level;
        int fnodecount;
        char type;

        ArrayList<Node> children = new ArrayList<Node>(); 

        public Node(int x){
            this.x = x;
            // this.rank = 0;
            this.type = 'f';
        }
    }



    static class DisjointUnionSets { 
        int[] rank, parent, ele_count; 
        int n; 
        
        // Constructor 
        public DisjointUnionSets(int n) 
        { 
            rank = new int[n]; 
            parent = new int[n];
            ele_count = new int[n];
            this.n = n; 
            makeSet(); 
        } 
        
        // Creates n sets with single item in each 
        void makeSet() 
        { 
            for (int i = 0; i < n; i++) { 
                // Initially, all elements are in 
                // their own set. 
                parent[i] = i; 
                ele_count[i] = 1;
            } 
        } 
        
        // Returns representative of x's set 
        int find(int x) 
        { 
            // Finds the representative of the set 
            // that x is an element of 
            if (parent[x] != x) { 
                // if x is not the parent of itself 
                // Then x is not the representative of 
                // his set,

                parent[x] = find(parent[x]); 
                
                // so we recursively call Find on its parent 
                // and move i's node directly under the 
                // representative of this set 
            } 
        
            return parent[x]; 
        }



        int setSize(int x){
            int par = find(x);
            return ele_count[par];
        }
        


        // Unites the set that includes x and the set 
        // that includes x 
        void union(int x, int y) 
        { 
            // Find representatives of two sets 
            int xRoot = find(x), yRoot = find(y); 
        
            // Elements are in the same set, no need 
            // to unite anything. 
            if (xRoot == yRoot) 
                return; 
        
            // If x's rank is less than y's rank 
            if (rank[xRoot] < rank[yRoot]){ 
        
                // Then move x under y  so that depth 
                // of tree remains less 
                parent[xRoot] = yRoot;
            
                ele_count[yRoot] = ele_count[yRoot] + ele_count[xRoot];
        
            // Else if y's rank is less than x's rank 
            }else if (rank[yRoot] < rank[xRoot]) {
        
                // Then move y under x so that depth of 
                // tree remains less 
                parent[yRoot] = xRoot; 
                ele_count[xRoot] = ele_count[yRoot] + ele_count[xRoot];
            
            } else // if ranks are the same 
            { 
                // Then move y under x (doesn't matter 
                // which one goes where) 
                parent[yRoot] = xRoot; 
        
                ele_count[xRoot] = ele_count[yRoot] + ele_count[xRoot];

                // And increment the result tree's 
                // rank by 1 
                rank[xRoot] = rank[xRoot] + 1; 
            } 
        } 
    } 


    static long power(long x, long y, long p) 
    { 
        // Initialize result 
        long res = 1l;      
        
        // Update x if it is more   
        // than or equal to p 
        x = x % p;  
    
        while (y > 0) 
        { 
            // If y is odd, multiply x 
            // with result 
            if((y & 1)==1) 
                res = (res * x) % p; 
    
            // y must be even now 
            // y = y / 2 
            y = y >> 1;  
            x = (x * x) % p;  
        } 
        return res; 
    } 



static int countGreater(int arr[], int n, int k) 
{ 
int l = 0; 
int r = n - 1; 

// Stores the index of the left most element 
// from the array which is greater than k 
int leftGreater = n; 

// Finds number of elements greater than k 
while (l <= r) { 
    int m = l + (r - l) / 2; 

    // If mid element is greater than 
    // k update leftGreater and r 
    if (arr[m] > k) { 
        leftGreater = m; 
        r = m - 1; 
    } 

    // If mid element is less than 
    // or equal to k update l 
    else
        l = m + 1; 
} 

// Return the count of elements greater than k 
return (n - leftGreater); 
} 



static int countLessEqual(int arr[], int n, int k) 
{ 
int l = 0; 
int r = n - 1; 

// Stores the index of the left most element 
// from the array which is greater than k 
int leftGreater = n; 

// Finds number of elements greater than k 
while (l <= r) { 
    int m = l + (r - l) / 2; 

    // If mid element is greater than 
    // k update leftGreater and r 
    if (arr[m] > k) { 
        leftGreater = m; 
        r = m - 1; 
    } 

    // If mid element is less than 
    // or equal to k update l 
    else
        l = m + 1; 
} 

// Return the count of elements greater than k 
return (leftGreater); 
} 


static int digiSum(long n){
    int sum = 0;
    while(n>0){
        sum+=(n%10);
        n/=10;
    }
    return sum;
}

public static void main(String args[]){
    try{
        // FastReader s = new FastReader();
        Scanner s = new Scanner(System.in);
        BufferedWriter w = new BufferedWriter(new OutputStreamWriter(System.out));
        int i, q, j, k, x, y, t, n, ind, count, diff, key, l, r, a, b, size, Pairer, max, sec_max, minglob, minloc, beg, end, begl, endr, up, down, maxele, minele, case_count;
        long ans, sqrt, sum, maxup, maxdown, m, X, Y, Q, MOD, num, lld;//, sum2;
        boolean flag, flag2;
        boolean l2r;

        float quo;

        double avg;
        char c, c2;
        String str;

        t = s.nextInt();
        for(; t>0; t--){
        
            num = s.nextLong();
            x = s.nextInt();

            y = digiSum(num);

            m = 10l; ans = 0l;
            while(y>x){
                // if(m==0)
                //     w.write(m+"\n");
                // else{
                lld = (num%m);
                lld = (lld/(m/10));
                if(lld!=0){
                    num = num + ((10)-lld)*(m/10);
                    ans += ((10)-lld)*(m/10);
                }
                y = digiSum(num);
                m = m*10;
            }

            // int arr[] = new int[n];

    
            // for(i=0; i<n-1; i++){
                w.write((ans)+" ");
            // }      
            w.write("\n");
                
        }
        w.close();
    }catch(Exception e){
        e.printStackTrace();
    }
}
}
                
                
//========================================### FAST IO ###=========================================//
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;
    } 
}
    //====================================================================================================//