#include <bits/stdc++.h>
using namespace std;
string solve1(string n) {
  string res(n.length(), '9');
  for (char c = '8'; c >= '0'; c--) {
    string t(n.length(), c);
    if (t >= n) res = t;
  }
  return res;
}
string solve2(string n) {
  string res = solve1(n);
  for (char a = '0'; a <= '9'; a++)
    for (char b = a + 1; b <= '9'; b++) {
      bool n_ok = true;
      for (int i = 0; i < n.length(); i++) {
        if (n[i] < b) {
          string t = n;
          if (t[i] < a)
            t[i] = a;
          else
            t[i] = b;
          for (int j = i + 1; j < n.length(); j++) t[j] = a;
          if (res > t) res = t;
        }
        if (n[i] != a && n[i] != b) {
          n_ok = false;
          break;
        }
      }
      if (n_ok) return n;
    }
  return res;
}
string solve() {
  string n;
  int k;
  cin >> n >> k;
  if (k == 1)
    return solve1(n);
  else
    return solve2(n);
}
int main() {
  int t;
  cin >> t;
  while (t--) cout << solve() << '\n';
  return 0;
}

### Prompt

Your challenge is to write 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;
string solve1(string n) {
  string res(n.length(), '9');
  for (char c = '8'; c >= '0'; c--) {
    string t(n.length(), c);
    if (t >= n) res = t;
  }
  return res;
}
string solve2(string n) {
  string res = solve1(n);
  for (char a = '0'; a <= '9'; a++)
    for (char b = a + 1; b <= '9'; b++) {
      bool n_ok = true;
      for (int i = 0; i < n.length(); i++) {
        if (n[i] < b) {
          string t = n;
          if (t[i] < a)
            t[i] = a;
          else
            t[i] = b;
          for (int j = i + 1; j < n.length(); j++) t[j] = a;
          if (res > t) res = t;
        }
        if (n[i] != a && n[i] != b) {
          n_ok = false;
          break;
        }
      }
      if (n_ok) return n;
    }
  return res;
}
string solve() {
  string n;
  int k;
  cin >> n >> k;
  if (k == 1)
    return solve1(n);
  else
    return solve2(n);
}
int main() {
  int t;
  cin >> t;
  while (t--) cout << solve() << '\n';
  return 0;
}

```