#include <bits/stdc++.h>
using namespace std;
bool ok(string s, int d) {
  sort(s.begin(), s.end());
  return unique(s.begin(), s.end()) - s.begin() <= d;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s;
    int d;
    cin >> s >> d;
    if (ok(s, d)) {
      cout << s << endl;
      continue;
    }
    for (int i = (int)s.size() - 1; i >= 0; i--) {
      for (s[i]++; s[i] <= '9'; s[i]++) {
        for (char j = '0'; j <= '9'; j++) {
          for (int k = i + 1; k < (int)s.size(); k++) s[k] = j;
          if (ok(s, d)) {
            cout << s << endl;
            goto next;
          }
        }
      }
      s[i] = '0';
    }
  next:;
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
bool ok(string s, int d) {
  sort(s.begin(), s.end());
  return unique(s.begin(), s.end()) - s.begin() <= d;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s;
    int d;
    cin >> s >> d;
    if (ok(s, d)) {
      cout << s << endl;
      continue;
    }
    for (int i = (int)s.size() - 1; i >= 0; i--) {
      for (s[i]++; s[i] <= '9'; s[i]++) {
        for (char j = '0'; j <= '9'; j++) {
          for (int k = i + 1; k < (int)s.size(); k++) s[k] = j;
          if (ok(s, d)) {
            cout << s << endl;
            goto next;
          }
        }
      }
      s[i] = '0';
    }
  next:;
  }
}

```