#include <bits/stdc++.h>
using namespace std;
int t, n, k;
int f() {
  int cnt = to_string(n).size();
  if (k == 1) {
    int ans = 0;
    for (int i = 0; i < cnt; i++) ans = ans * 10 + 1;
    for (int i = 0; i < 10; i++)
      if (i * ans >= n) return i * ans;
  }
  int ans = 2e9;
  int s = to_string(n)[0] - '0';
  for (int i = 0; i <= 1 << cnt; i++)
    for (int j = s; j < s + 1; j++)
      for (int k = 0, x, m, cur; k < 10; k++) {
        for (m = i, x = cur = 0; x < cnt; x++, m /= 2)
          cur = cur * 10 + (m % 2 ? j : k);
        if (cur >= n) ans = min(cur, ans);
      }
  return ans;
}
signed main() {
  for (cin >> t; t--;) {
    cin >> n >> k;
    cout << f() << "\n";
  }
}

### Prompt

Create a solution in Cpp for 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;
int t, n, k;
int f() {
  int cnt = to_string(n).size();
  if (k == 1) {
    int ans = 0;
    for (int i = 0; i < cnt; i++) ans = ans * 10 + 1;
    for (int i = 0; i < 10; i++)
      if (i * ans >= n) return i * ans;
  }
  int ans = 2e9;
  int s = to_string(n)[0] - '0';
  for (int i = 0; i <= 1 << cnt; i++)
    for (int j = s; j < s + 1; j++)
      for (int k = 0, x, m, cur; k < 10; k++) {
        for (m = i, x = cur = 0; x < cnt; x++, m /= 2)
          cur = cur * 10 + (m % 2 ? j : k);
        if (cur >= n) ans = min(cur, ans);
      }
  return ans;
}
signed main() {
  for (cin >> t; t--;) {
    cin >> n >> k;
    cout << f() << "\n";
  }
}

```