#include <bits/stdc++.h>
using namespace std;
void solve() {
  string N;
  int K;
  cin >> N >> K;
  while (true) {
    set<char> st;
    for (int i = 0; i < N.size(); i++) st.insert(N[i]);
    if (st.size() <= K) {
      cout << N << endl;
      return;
    }
    st.clear();
    int nw = 0;
    for (;; nw++) {
      st.insert(N[nw]);
      if (st.size() > K) {
        while (N[nw] == '9') nw -= 1;
        N[nw]++;
        for (int i = nw + 1; i < N.size(); i++) N[i] = '0';
        break;
      }
    }
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) solve();
}

### Prompt

Generate 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;
void solve() {
  string N;
  int K;
  cin >> N >> K;
  while (true) {
    set<char> st;
    for (int i = 0; i < N.size(); i++) st.insert(N[i]);
    if (st.size() <= K) {
      cout << N << endl;
      return;
    }
    st.clear();
    int nw = 0;
    for (;; nw++) {
      st.insert(N[nw]);
      if (st.size() > K) {
        while (N[nw] == '9') nw -= 1;
        N[nw]++;
        for (int i = nw + 1; i < N.size(); i++) N[i] = '0';
        break;
      }
    }
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) solve();
}

```