#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    string str;
    int k;
    cin >> str >> k;
    while (true) {
      set<char> st;
      for (int i = 0; i < str.size(); i++) st.insert(str[i]);
      if (st.size() <= k) break;
      st.clear();
      for (int i = 0; i < str.size(); i++) {
        st.insert(str[i]);
        if (st.size() > k) {
          int pos = i;
          while (str[pos] == '9') pos--;
          str[pos] += 1;
          for (int j = pos + 1; j < str.size(); j++) {
            str[j] = '0';
          }
          break;
        }
      }
    }
    cout << str << endl;
  }
  return 0;
}

### Prompt

Generate 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;
signed main() {
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    string str;
    int k;
    cin >> str >> k;
    while (true) {
      set<char> st;
      for (int i = 0; i < str.size(); i++) st.insert(str[i]);
      if (st.size() <= k) break;
      st.clear();
      for (int i = 0; i < str.size(); i++) {
        st.insert(str[i]);
        if (st.size() > k) {
          int pos = i;
          while (str[pos] == '9') pos--;
          str[pos] += 1;
          for (int j = pos + 1; j < str.size(); j++) {
            str[j] = '0';
          }
          break;
        }
      }
    }
    cout << str << endl;
  }
  return 0;
}

```