#include <bits/stdc++.h>
using namespace std;
string solve1(string s) {
  string temp1 = "";
  string temp2 = "";
  if (s[0] != 9) {
    for (int i = 0; i < s.length(); i++) {
      temp1 += s[0];
      temp2 += s[0] + 1;
    }
    if (temp1 >= s) {
      return temp1;
    } else {
      return temp2;
    }
  } else {
    for (int i = 0; i < s.length(); i++) {
      temp1 += s[0];
    }
    return temp1;
  }
}
void solve2(string s) {
  string res = solve1(s);
  for (char a = '0'; a <= '9'; a++) {
    for (char b = '1'; b <= '9'; b++) {
      bool s_ok = true;
      for (int i = 0; i < s.length(); i++) {
        if (s[i] < b) {
          string t = s;
          if (t[i] < a) {
            t[i] = a;
          } else {
            t[i] = b;
          }
          for (int j = i + 1; j < s.length(); j++) {
            t[j] = a;
          }
          if (res > t) {
            res = t;
          }
        }
        if (s[i] != a && s[i] != b) {
          s_ok = false;
          break;
        }
      }
      if (s_ok) {
        cout << s << endl;
        return;
      }
    }
  }
  cout << res << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long int n, k;
    cin >> n >> k;
    string s = "";
    long long int temp = n;
    while (temp != 0) {
      s.push_back((temp % 10) + '0');
      temp /= 10;
    }
    reverse(s.begin(), s.end());
    if (k == 1) {
      cout << solve1(s) << endl;
    } else {
      solve2(s);
    }
  }
  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 s) {
  string temp1 = "";
  string temp2 = "";
  if (s[0] != 9) {
    for (int i = 0; i < s.length(); i++) {
      temp1 += s[0];
      temp2 += s[0] + 1;
    }
    if (temp1 >= s) {
      return temp1;
    } else {
      return temp2;
    }
  } else {
    for (int i = 0; i < s.length(); i++) {
      temp1 += s[0];
    }
    return temp1;
  }
}
void solve2(string s) {
  string res = solve1(s);
  for (char a = '0'; a <= '9'; a++) {
    for (char b = '1'; b <= '9'; b++) {
      bool s_ok = true;
      for (int i = 0; i < s.length(); i++) {
        if (s[i] < b) {
          string t = s;
          if (t[i] < a) {
            t[i] = a;
          } else {
            t[i] = b;
          }
          for (int j = i + 1; j < s.length(); j++) {
            t[j] = a;
          }
          if (res > t) {
            res = t;
          }
        }
        if (s[i] != a && s[i] != b) {
          s_ok = false;
          break;
        }
      }
      if (s_ok) {
        cout << s << endl;
        return;
      }
    }
  }
  cout << res << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long int n, k;
    cin >> n >> k;
    string s = "";
    long long int temp = n;
    while (temp != 0) {
      s.push_back((temp % 10) + '0');
      temp /= 10;
    }
    reverse(s.begin(), s.end());
    if (k == 1) {
      cout << solve1(s) << endl;
    } else {
      solve2(s);
    }
  }
  return 0;
}

```