#include<bits/stdc++.h>
#include <string>
using namespace std;
typedef long long ll;
#define endl '\n'

const ll NO=3e18;
//string str;

ll dp[25][25][2];
ll mi[20]={1};

ll inline fun1(ll a,ll b ){
    return (a/mi[b])%10;
}

ll inline fun2(ll a,ll b){
    return a*mi[b];
}

ll fun3(ll a,ll b){
    int cnt=0;
    while(a||b){
        if(a%10!=b%10)cnt++;
        a/=10;
        b/=10;
    }
    return cnt-1;
}


void solve(){
    for(int i=0;i<=21;i++){
        for(int j=0;j<=21;j++){
            dp[i][j][1]=dp[i][j][0]=NO;
        }
    }
    ll x;
    ll k;

    cin>>x>>k;

	//str="";

    if(k==0){
        ll y=1;
        while(x%10==9){
            y*=10;
            x/=10;
        }
		//str+=to_string(y);
		//str+='\n';
		printf("%lld\n",y);
        return;

    }


    int len=0;
    while(x%10==0){
        len++;
        x/=10;
    }

    for(int i=0;i<=20;i++){
        dp[i][0][0]=x;
    }
    for(int i=0;i<=18;i++){
        for(int j=0;j<=18;j++){
            //if(i<=7)cout<<"dp["<<i<<"]["<<j<<"]="<<(dp[i][j][0]==NO?-1:dp[i][j][0]-x)<<" "<<(dp[i][j][1]==NO?-1:dp[i][j][1]-x)<<endl;
            //if(i!=0)dp[i][j][0]=min(dp[i][j][0],dp[i-1][j][0]);
            //if(i!=0)dp[i][j][0]=min(dp[i][j][0],dp[i-1][j][1]);
            if(dp[i][j][1]!=NO){
                int tem=fun1(dp[i][j][1],i);
                if(tem!=0){
                    tem=10-tem;
                    ll b=dp[i][j][1]+fun2(tem,i);
                    int ch=fun3(dp[i][j][1],b);
                    dp[i+ch][j+ch][1]=min(dp[i+ch][j+ch][1],b);
					//cout<<"    dp["<<i+ch<<"]["<<j+ch<<"][1]="<<min(dp[i+ch][j+ch][1],b)-x<<endl;
            }
                }
                dp[i+1][j][0]=min(dp[i+1][j][0],dp[i][j][1]);
                
            if(dp[i][j][0]!=NO){
                int tem=fun1(dp[i][j][0],i);
                if(tem!=0){
                    tem=10-tem;
                    ll b=dp[i][j][0]+fun2(tem,i);
                    int ch=fun3(dp[i][j][0],b);
                    dp[i+ch][j+ch][1]=min(dp[i+ch][j+ch][1],b);
					//cout<<"    dp["<<i+ch<<"]["<<j+ch<<"][1]="<<min(dp[i+ch][j+ch][1],b)-x<<endl;
                }
                dp[i+1][j][0]=min(dp[i+1][j][0],dp[i][j][0]);
                
            }
            
        }
    }
    ll miny=NO;
    for(int i=0;i<19;i++){
        
        miny=min(miny,dp[i][k][0]);
    }
    //cout<<miny<<endl;
    for(int i=0;i<19;i++){
        miny=min(miny,dp[i][k][1]);
        //if(dp[i][k][1]<=0)cout<<i<<endl;
    }
    //cout<<miny<<endl;
	//str+=to_string(miny-x);
	printf("%lld",miny-x);
    while(len--){
        //str+='0';
		printf("0");
    }
	printf("\n");
    //str+='\n';
    //cout<<min(dp[18][k]-x,dp[17][k]-x)<<endl;
}


int main(){
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    int tt=1;
    cin>>tt;

    for(int i=1;i<=19;i++){
        mi[i]=mi[i-1]*10;
    }
    while(tt--){
        solve();
		//cout<<str;
    }
    

    return 0;
}