#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const ll INF = 4e18;
const int maxn = 50;
ll dp[maxn][maxn][2];//dp[i][j][0/1]表示从低位开始到第i位，以及进位过j次且当前位是否向前进位情况下的最小y值
ll p_10[maxn];

void solve()
{
    ll x,k;
    cin>>x>>k;
    if(k == 0){
        //k == 0特判
        ll ans = 1;
        while(x%10 == 9){
            x/=10;
            ans*=10;
        }
        cout<<ans<<"\n";
        return;
    }
    //提前对x末尾的0做出预处理，防止爆long long 
    int cnt_0 = 0;
    while(x%10 == 0){
        cnt_0++;
        x/=10;
    }
    //x存入数组
    vector<int>num(50,0);
    int pos = 0;
    while(x){
        num[++pos] = x%10;
        x/=10;
    }
    for(int i=0;i<=20;i++)
        for(int j=0;j<=20;j++)
            for(int k=0;k<=1;k++){
                dp[i][j][k] = INF;
            }
    dp[0][0][0] = 0;
    for(int i=1;i<=18;i++)
        for(int j=0;j<=i;j++)
        {
            if(num[i] == 9){
                dp[i][j][0] = min(dp[i][j][0],dp[i-1][j][0]);
                if(j>=1)dp[i][j][1] = min(dp[i][j][1],dp[i-1][j-1][1]);
                if(j>=1)dp[i][j][1] = min(dp[i][j][1],dp[i-1][j-1][0] + p_10[i-1]);
            }
            else{
                dp[i][j][0] = min(dp[i][j][0],min(dp[i-1][j][0],dp[i-1][j][1]));
                if(j>=1)dp[i][j][1] = min(dp[i][j][1],dp[i-1][j-1][0] + p_10[i-1]*(10 - num[i]));
                if(j>=1)dp[i][j][1] = min(dp[i][j][1],dp[i-1][j-1][1] + p_10[i-1]*(9-num[i]));
            }
        }
    // cerr<<dp[2][1][1]<<"\n";
    cout<<min(dp[18][k][0],dp[18][k][1]);
    while(cnt_0--)cout<<"0";
    cout<<"\n";
    return;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin>>t;
    p_10[0] = 1;
    for(int i=1;i<=18;i++){
        p_10[i] = p_10[i-1]*10;
    }
    while(t--){
        solve();
    }
    return 0;
}