#include <bits/stdc++.h>
using namespace std;
long long k, t, p[100010], d;
long long minn, maxn, n;
int dig(int x) {
  int ans = 0;
  while (x) {
    ans++;
    p[ans] = x % 10;
    maxn = max(maxn, p[ans]);
    x = x / 10;
  }
  return ans;
}
void dfs(int id, int x, long long sum) {
  if (id == 1) {
    if (sum >= n) minn = min(minn, sum);
    return;
  }
  dfs(id - 1, x, sum * 10 + x);
  dfs(id - 1, x, sum * 10 + p[d]);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> t;
  while (t--) {
    cin >> n >> k;
    minn = 1e11;
    maxn = 0;
    d = dig(n);
    if (k == 1) {
      long long s;
      for (int i = 1; i <= 9; i++) {
        s = 0;
        for (int j = 1; j <= d; j++) {
          s = s * 10 + i;
        }
        if (s >= n) minn = min(minn, s);
      }
      cout << minn << "\n";
      continue;
    }
    for (int i = 0; i <= 9; i++) dfs(d, i, p[d]);
    cout << minn << "\n";
  }
}

### Prompt

Your task is to create 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;
long long k, t, p[100010], d;
long long minn, maxn, n;
int dig(int x) {
  int ans = 0;
  while (x) {
    ans++;
    p[ans] = x % 10;
    maxn = max(maxn, p[ans]);
    x = x / 10;
  }
  return ans;
}
void dfs(int id, int x, long long sum) {
  if (id == 1) {
    if (sum >= n) minn = min(minn, sum);
    return;
  }
  dfs(id - 1, x, sum * 10 + x);
  dfs(id - 1, x, sum * 10 + p[d]);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> t;
  while (t--) {
    cin >> n >> k;
    minn = 1e11;
    maxn = 0;
    d = dig(n);
    if (k == 1) {
      long long s;
      for (int i = 1; i <= 9; i++) {
        s = 0;
        for (int j = 1; j <= d; j++) {
          s = s * 10 + i;
        }
        if (s >= n) minn = min(minn, s);
      }
      cout << minn << "\n";
      continue;
    }
    for (int i = 0; i <= 9; i++) dfs(d, i, p[d]);
    cout << minn << "\n";
  }
}

```