#include <bits/stdc++.h>
using namespace std;
template <typename T>
void get_vector(T &a) {
  for (auto &e : a) cin >> e;
}
template <typename T>
void put_vector(T a) {
  for (auto e : a) cout << e << " ";
  cout << endl;
}
void solve() {
  string s;
  int k;
  cin >> s >> k;
  while (1) {
    set<char> cnt;
    for (auto e : s) cnt.insert(e);
    if (cnt.size() <= k) {
      cout << s << endl;
      return;
    }
    cnt.clear();
    int ptr = 0;
    while (1) {
      cnt.insert(s[ptr]);
      if (cnt.size() > k) {
        while (s[ptr] == '9') ptr--;
        s[ptr]++;
        for (int i = ptr + 1; i < s.size(); i++) s[i] = '0';
        break;
      }
      ptr++;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T;
  cin >> T;
  while (T--) solve();
  return 0;
}

### Prompt

Please create a solution in CPP 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;
template <typename T>
void get_vector(T &a) {
  for (auto &e : a) cin >> e;
}
template <typename T>
void put_vector(T a) {
  for (auto e : a) cout << e << " ";
  cout << endl;
}
void solve() {
  string s;
  int k;
  cin >> s >> k;
  while (1) {
    set<char> cnt;
    for (auto e : s) cnt.insert(e);
    if (cnt.size() <= k) {
      cout << s << endl;
      return;
    }
    cnt.clear();
    int ptr = 0;
    while (1) {
      cnt.insert(s[ptr]);
      if (cnt.size() > k) {
        while (s[ptr] == '9') ptr--;
        s[ptr]++;
        for (int i = ptr + 1; i < s.size(); i++) s[i] = '0';
        break;
      }
      ptr++;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T;
  cin >> T;
  while (T--) solve();
  return 0;
}

```