#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n, k;
    cin >> n >> k;
    string N = to_string(n);
    string ans = "99999999999999";
    set<char> S;
    for (auto c : N) {
      S.insert(c);
    }
    if ((int)S.size() <= k) {
      cout << N << "\n";
      continue;
    }
    for (int d = 0; d < (int)N.size(); d++) {
      string base = N.substr(0, d);
      for (int v = N[d] - '0' + 1; v <= 9; v++) {
        string s = base + to_string(v);
        set<char> have;
        for (int i = 0; i < d; i++) {
          have.insert(N[i]);
        }
        have.insert((char)(v + '0'));
        if ((int)have.size() > k) {
          continue;
        }
        if ((int)have.size() < k) {
          have.insert('0');
        }
        while (s.size() < N.size()) {
          s += *have.begin();
        }
        assert(s >= N);
        ans = min(ans, s);
      }
    }
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

It is a simplified version of problem F2. The difference between them is the constraints (F1: k ≤ 2, F2: k ≤ 10).

You are given an integer n. Find the minimum integer x such that x ≥ n and the number x is k-beautiful.

A number is called k-beautiful if its decimal representation having no leading zeroes contains no more than k different digits. E.g. if k = 2, the numbers 3434443, 55550, 777 and 21 are k-beautiful whereas the numbers 120, 445435 and 998244353 are not.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

Each test case consists of one line containing two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2).

Output

For each test case output on a separate line x — the minimum k-beautiful integer such that x ≥ n.

Example

Input


4
1 1
221 2
177890 2
998244353 1


Output


1
221
181111
999999999

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n, k;
    cin >> n >> k;
    string N = to_string(n);
    string ans = "99999999999999";
    set<char> S;
    for (auto c : N) {
      S.insert(c);
    }
    if ((int)S.size() <= k) {
      cout << N << "\n";
      continue;
    }
    for (int d = 0; d < (int)N.size(); d++) {
      string base = N.substr(0, d);
      for (int v = N[d] - '0' + 1; v <= 9; v++) {
        string s = base + to_string(v);
        set<char> have;
        for (int i = 0; i < d; i++) {
          have.insert(N[i]);
        }
        have.insert((char)(v + '0'));
        if ((int)have.size() > k) {
          continue;
        }
        if ((int)have.size() < k) {
          have.insert('0');
        }
        while (s.size() < N.size()) {
          s += *have.begin();
        }
        assert(s >= N);
        ans = min(ans, s);
      }
    }
    cout << ans << "\n";
  }
  return 0;
}

```