#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t;
  cin >> t;
  for (int test = 0; test < t; test++) {
    string s;
    int n;
    cin >> s >> n;
    set<char> st;
    map<char, int> fst;
    int p = 0, lst = -1;
    while (p < int(s.size()) and int(st.size()) <= n) {
      if (st.count(s[p]) == 0) {
        fst[s[p]] = p;
        st.insert(s[p]);
        if (st.size() == n) {
          lst = p;
        }
      }
      p++;
    }
    if (int(st.size()) <= n) {
      cout << s << '\n';
    } else {
      p--;
      st.erase(s[p]);
      if ((*st.rbegin()) > s[p]) {
        s[p] = (*st.upper_bound(s[p]));
        for (p = p + 1; p < s.size(); p++) {
          s[p] = (*st.begin());
        }
      } else {
        int flag = 0;
        while (!flag and p > lst + 1) {
          p--;
          if ((*st.rbegin()) > s[p]) {
            s[p] = (*st.upper_bound(s[p]));
            for (int pp = p + 1; pp < s.size(); pp++) {
              s[pp] = (*st.begin());
            }
            flag = 1;
          }
        }
        if (flag == 0) {
          p = lst;
          st.erase(s[p]);
          if (st.count(s[p] + 1) > 0) {
            s[p]++;
            for (int pp = p + 1; pp < s.size(); pp++) {
              s[pp] = '0';
            }
          } else {
            s[p] += 1;
            st.insert(s[p]);
            for (p = p + 1; p < s.size(); p++) {
              s[p] = (*st.begin());
            }
          }
        }
      }
      cout << s << '\n';
    }
  }
  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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t;
  cin >> t;
  for (int test = 0; test < t; test++) {
    string s;
    int n;
    cin >> s >> n;
    set<char> st;
    map<char, int> fst;
    int p = 0, lst = -1;
    while (p < int(s.size()) and int(st.size()) <= n) {
      if (st.count(s[p]) == 0) {
        fst[s[p]] = p;
        st.insert(s[p]);
        if (st.size() == n) {
          lst = p;
        }
      }
      p++;
    }
    if (int(st.size()) <= n) {
      cout << s << '\n';
    } else {
      p--;
      st.erase(s[p]);
      if ((*st.rbegin()) > s[p]) {
        s[p] = (*st.upper_bound(s[p]));
        for (p = p + 1; p < s.size(); p++) {
          s[p] = (*st.begin());
        }
      } else {
        int flag = 0;
        while (!flag and p > lst + 1) {
          p--;
          if ((*st.rbegin()) > s[p]) {
            s[p] = (*st.upper_bound(s[p]));
            for (int pp = p + 1; pp < s.size(); pp++) {
              s[pp] = (*st.begin());
            }
            flag = 1;
          }
        }
        if (flag == 0) {
          p = lst;
          st.erase(s[p]);
          if (st.count(s[p] + 1) > 0) {
            s[p]++;
            for (int pp = p + 1; pp < s.size(); pp++) {
              s[pp] = '0';
            }
          } else {
            s[p] += 1;
            st.insert(s[p]);
            for (p = p + 1; p < s.size(); p++) {
              s[p] = (*st.begin());
            }
          }
        }
      }
      cout << s << '\n';
    }
  }
  return 0;
}

```