#include <bits/stdc++.h>
using namespace std;
using ll = long long;
string n;
void solve1() {
  ll len = n.size();
  for (char c = '1'; c <= '9'; ++c) {
    string ans = "";
    for (ll i = 1; i <= len; ++i) ans += c;
    if (ans >= n) {
      cout << ans << '\n';
      return;
    }
  }
}
void solve2() {
  ll len = n.size();
  string res;
  for (char c = '1'; c <= '9'; ++c) {
    string ans = "";
    for (ll i = 1; i <= len; ++i) ans += c;
    if (ans >= n) {
      res = ans;
      break;
    }
  }
  for (char a = '0'; a <= '9'; ++a) {
    for (char b = a + 1; b <= '9'; ++b) {
      bool ok = 1;
      for (ll i = 0; i < len; ++i) {
        if (n[i] < b) {
          string ans = n;
          if (n[i] < a)
            ans[i] = a;
          else
            ans[i] = b;
          for (ll j = i + 1; j < len; j++) ans[j] = a;
          if (ans < res) res = ans;
        }
        if (n[i] != a && n[i] != b) {
          ok = 0;
          break;
        }
      }
      if (ok) {
        cout << n << '\n';
        return;
      }
    }
  }
  cout << res << '\n';
  return;
}
void solve() {
  ll k;
  cin >> n;
  cin >> k;
  if (k == 1)
    solve1();
  else
    solve2();
}
signed main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ll t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Please formulate 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;
using ll = long long;
string n;
void solve1() {
  ll len = n.size();
  for (char c = '1'; c <= '9'; ++c) {
    string ans = "";
    for (ll i = 1; i <= len; ++i) ans += c;
    if (ans >= n) {
      cout << ans << '\n';
      return;
    }
  }
}
void solve2() {
  ll len = n.size();
  string res;
  for (char c = '1'; c <= '9'; ++c) {
    string ans = "";
    for (ll i = 1; i <= len; ++i) ans += c;
    if (ans >= n) {
      res = ans;
      break;
    }
  }
  for (char a = '0'; a <= '9'; ++a) {
    for (char b = a + 1; b <= '9'; ++b) {
      bool ok = 1;
      for (ll i = 0; i < len; ++i) {
        if (n[i] < b) {
          string ans = n;
          if (n[i] < a)
            ans[i] = a;
          else
            ans[i] = b;
          for (ll j = i + 1; j < len; j++) ans[j] = a;
          if (ans < res) res = ans;
        }
        if (n[i] != a && n[i] != b) {
          ok = 0;
          break;
        }
      }
      if (ok) {
        cout << n << '\n';
        return;
      }
    }
  }
  cout << res << '\n';
  return;
}
void solve() {
  ll k;
  cin >> n;
  cin >> k;
  if (k == 1)
    solve1();
  else
    solve2();
}
signed main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ll t;
  cin >> t;
  while (t--) solve();
}

```