#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 7;
set<long long> s1;
set<long long> s2;
long long p[20];
void gen() {
  p[0] = 1;
  for (long long i = 1; i <= 15; i++) p[i] = p[i - 1] * 10;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  if (k == 1)
    cout << *s1.lower_bound(n) << "\n";
  else
    cout << *s2.lower_bound(n) << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  gen();
  for (long long i = 1; i <= 9; i++) {
    long long liczba = 0;
    for (long long j = 1; j <= 10; j++) {
      liczba *= 10;
      liczba += i;
      s1.insert(liczba);
    }
  }
  for (long long i = 0; i <= 9; i++)
    for (long long j = i + 1; j <= 9; j++)
      for (long long l = 0; l <= 10; l++)
        for (long long mask = 0; mask <= (1 << l) - 1; mask++) {
          long long liczba = 0;
          for (long long m = 0; m <= l - 1; m++) {
            if (mask & (1 << m))
              liczba += p[m] * i;
            else
              liczba += p[m] * j;
          }
          s2.insert(liczba);
        }
  int t;
  cin >> t;
  while (t--) solve();
}

### 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;
const int INF = 1e9 + 7;
set<long long> s1;
set<long long> s2;
long long p[20];
void gen() {
  p[0] = 1;
  for (long long i = 1; i <= 15; i++) p[i] = p[i - 1] * 10;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  if (k == 1)
    cout << *s1.lower_bound(n) << "\n";
  else
    cout << *s2.lower_bound(n) << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  gen();
  for (long long i = 1; i <= 9; i++) {
    long long liczba = 0;
    for (long long j = 1; j <= 10; j++) {
      liczba *= 10;
      liczba += i;
      s1.insert(liczba);
    }
  }
  for (long long i = 0; i <= 9; i++)
    for (long long j = i + 1; j <= 9; j++)
      for (long long l = 0; l <= 10; l++)
        for (long long mask = 0; mask <= (1 << l) - 1; mask++) {
          long long liczba = 0;
          for (long long m = 0; m <= l - 1; m++) {
            if (mask & (1 << m))
              liczba += p[m] * i;
            else
              liczba += p[m] * j;
          }
          s2.insert(liczba);
        }
  int t;
  cin >> t;
  while (t--) solve();
}

```