#include <bits/stdc++.h>
#define endl "\n"
using namespace std;
typedef unsigned long long LL;
typedef pair<LL, LL> PII;
const LL N = 19, INF = 0x3f3f3f3f, mod = 1e9 + 7;
LL pow10[60];

LL f(LL x, LL k)
{
    if (k == 1) return x % 10;
    LL res = pow10[k - 1];
    return x / res % 10;
}

void solve()
{
    LL x, k;
    cin >> x >> k;

    LL cnt = 0;
    while (x % 10 == 0) x /= 10, cnt++; 

    if (k == 0)
    {
        for (LL i = 0; i <= N; i++)
            if (x / pow10[i] % 10 != 9)
            {
                cout << pow10[i] << endl;
                while (cnt--)   cout << "0";
                return;
            }
    }
    
    LL ans = 0;
    while (k)
    {
        if (f(x, k + 1) != 9)
        {
            LL res = pow10[k];
            ans += res - x % res;
            break;
        }
        else
        {
            LL st = k, en = k;
            while (st < 20 && f(x, st) == 9)   st++;
            while (en && f(x, en) == 9)   en--;
            st--, en++;
            
            if (st - en + 1 >= k)
            {
                ans += pow10[st - k + 1 - 1];
                break;
            }
            else
            {
                LL posx = en - 1;
                while (f(x, posx) == 0) posx--;
                if (st - posx + 1 <= k)
                {
                    ans += (pow10[st - posx + 1] - x % pow10[st] / pow10[posx - 1]) * pow10[posx - 1];
                    k -= st - posx + 1;
                }
                else
                {
                    ans += pow10[en];
                    k -= st - en + 1;
                }
            }
        }
    }
    cout << ans;
    while (cnt--)   cout << "0";
    cout << endl;
}

signed main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);

    pow10[0] = 1;
    for (LL i = 1; i <= N; i++)
        pow10[i] = pow10[i - 1] * 10;

    LL T = 1;
    cin >> T;
    while (T--)
        solve();
    return 0;
}