#include <bits/stdc++.h>
using namespace std;
int t;
int k;
string s;
string solve() {
  while (1) {
    set<char> q;
    for (int i = 0; i < s.size(); i++) q.insert(s[i]);
    if (q.size() <= k) return s;
    q.clear();
    for (int pre = 0; pre < s.size(); pre++) {
      q.insert(s[pre]);
      if (q.size() > k) {
        while (s[pre] == '9') pre--;
        s[pre]++;
        for (int i = pre + 1; i < s.size(); i++) {
          s[i] = '0';
        }
        break;
      }
    }
  }
}
int main() {
  cin >> t;
  while (t--) {
    cin >> s >> k;
    cout << solve() << endl;
  }
  return 0;
}

### Prompt

Develop a solution in CPP 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;
int t;
int k;
string s;
string solve() {
  while (1) {
    set<char> q;
    for (int i = 0; i < s.size(); i++) q.insert(s[i]);
    if (q.size() <= k) return s;
    q.clear();
    for (int pre = 0; pre < s.size(); pre++) {
      q.insert(s[pre]);
      if (q.size() > k) {
        while (s[pre] == '9') pre--;
        s[pre]++;
        for (int i = pre + 1; i < s.size(); i++) {
          s[i] = '0';
        }
        break;
      }
    }
  }
}
int main() {
  cin >> t;
  while (t--) {
    cin >> s >> k;
    cout << solve() << endl;
  }
  return 0;
}

```