#pragma GCC optimize("-Ofast","-ffast-math","-funroll-all-loops")
#include <bits/stdc++.h>
using namespace std;
#define ll __int128
#define db double
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define mod 998244353
#define vi vector<int>
const int maxn = 2000005;
const int S = 37;
template<typename T> bool chkmin(T &a, T b) {
    return (b < a) ? (a = b, 1) : 0;
}

template<typename T> bool chkmax(T &a, T b) {
    return (b > a) ? (a = b, 1) : 0;
}
ll dp[S + 2][3][S]; // next undet: i; j many < 
int dig[S];
int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
        ll x;
        int k;
        long long inp;
        cin >> inp >> k;
        x = inp;
        for (int j = 0; j < S; j++) {
            dig[j] = x % 10;
            x /= 10;
        }
        ll inf = 3e18;
        inf = inf * inf;
        for (int i = 0; i < S + 2; i++)
            for (int j = 0; j < 3; j++)
                for (int k = 0; k < S; k++)
                    dp[i][j][k] = inf;
        dp[0][1][0] = 0;
        ll pw = 1;
        for (int i = 0; i < S; i++) {
            vi ids;
            for (int w = 0; w < 10; w++) 
                if (w == dig[i] || w == 0 || w == dig[i] + 1) ids.pb(w);
            for (int u = 0; u < 3; u++)
                for (int v = 0; v <= 18; v++) {
                    if (dp[i][u][v] > inf) continue;
                    for (auto w : ids) {
                        // if (i == S - 1 && w >= 4) continue;
                        ll cc = dp[i][u][v] + w * pw;
                        int eu = 0, ev = v;
                        if (w > dig[i]) eu = 2;
                        else if (w == dig[i]) eu = u;
                        else eu = 0;
                        if (!eu) ev += 1;
                        chkmin(dp[i + 1][eu][ev], cc);
                    }
                }
            pw = pw * 10;
        }
        ll ans = dp[S][2][k] - inp;
        vi otp;
        while (ans) {
            otp.pb(ans % 10);
            ans /= 10;
        }
        reverse(otp.begin(), otp.end());
        for (auto v : otp) printf("%d", v);
        printf("\n");
        // cout << ans << '\n';
        // cout << dp[S][2][k] - stx << '\n';
    }
    return 0;
}
/*
4
12345678 0
12345678 5
12345678 18
990099 5
*/