#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 5e5 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-6;
int n, k;
long long sum(long long first) {
  long long ans = 0;
  while (first) {
    if (first & 1) ans++;
    first >>= 1;
  }
  return ans;
}
long long calc(long long first) {
  long long p = 0;
  while (first) {
    p |= (1 << (first % 10));
    first /= 10;
  }
  return sum(p);
}
void solve() {
  cin >> n >> k;
  while (calc(n) > k) {
    long long p = n;
    long long r = 1;
    while (calc(p) > k) {
      p /= 10;
      r *= 10;
    }
    r /= 10;
    n = ((n / r) + 1) * r;
  }
  cout << n << "\n";
}
signed main() {
  ios::sync_with_stdio(false);
  int test = 1;
  cin >> test;
  while (test--) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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;
const int MAX_N = 5e5 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-6;
int n, k;
long long sum(long long first) {
  long long ans = 0;
  while (first) {
    if (first & 1) ans++;
    first >>= 1;
  }
  return ans;
}
long long calc(long long first) {
  long long p = 0;
  while (first) {
    p |= (1 << (first % 10));
    first /= 10;
  }
  return sum(p);
}
void solve() {
  cin >> n >> k;
  while (calc(n) > k) {
    long long p = n;
    long long r = 1;
    while (calc(p) > k) {
      p /= 10;
      r *= 10;
    }
    r /= 10;
    n = ((n / r) + 1) * r;
  }
  cout << n << "\n";
}
signed main() {
  ios::sync_with_stdio(false);
  int test = 1;
  cin >> test;
  while (test--) {
    solve();
  }
  return 0;
}

```