#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 7;
const double eps = 1e-6;
const int mod = 1e9 + 7;
int a[N];
int digits(int n) {
  bool f[10] = {0};
  while (n) {
    f[n % 10] = 1;
    n /= 10;
  }
  int cnt = 0;
  for (int i = 0; i < 10; i++) cnt += f[i];
  return cnt;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k, t;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    while (digits(n) > k) {
      int r = 1;
      int c = n;
      while (digits(c) > k) {
        c /= 10;
        r *= 10;
      }
      r /= 10;
      n = ((n / r) + 1) * r;
    }
    cout << n << endl;
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
const int N = 1e6 + 7;
const double eps = 1e-6;
const int mod = 1e9 + 7;
int a[N];
int digits(int n) {
  bool f[10] = {0};
  while (n) {
    f[n % 10] = 1;
    n /= 10;
  }
  int cnt = 0;
  for (int i = 0; i < 10; i++) cnt += f[i];
  return cnt;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k, t;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    while (digits(n) > k) {
      int r = 1;
      int c = n;
      while (digits(c) > k) {
        c /= 10;
        r *= 10;
      }
      r /= 10;
      n = ((n / r) + 1) * r;
    }
    cout << n << endl;
  }
  return 0;
}

```