#include <bits/stdc++.h>
using namespace std;
set<int> used;
int k;
string digs;
int pre;
int go(int rem, bool match) {
  if (rem == 0) return pre;
  int p = digs.length() - rem;
  int tried = 0;
  for (int i = 0; i < 10; i++) {
    if (match and i < (int)digs[p] - '0') continue;
    if (used.size() == k and used.count(i) == 0) continue;
    if (tried >= 2) continue;
    tried++;
    pre = 10 * pre, pre += i;
    bool curin = used.count(i);
    if (!curin) used.insert(i);
    int res = go(rem - 1, (match and i == digs[p] - '0'));
    if (!curin) used.erase(i);
    pre /= 10;
    if (res != -1) {
      return res;
    }
  }
  return -1;
}
void solve() {
  int n;
  cin >> n >> k;
  used.clear();
  pre = 0;
  digs = to_string(n);
  int res = go(digs.length(), true);
  cout << res << endl;
}
int main() {
  int t;
  cin >> t;
  while (t) {
    solve();
    t--;
  }
}

### 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;
set<int> used;
int k;
string digs;
int pre;
int go(int rem, bool match) {
  if (rem == 0) return pre;
  int p = digs.length() - rem;
  int tried = 0;
  for (int i = 0; i < 10; i++) {
    if (match and i < (int)digs[p] - '0') continue;
    if (used.size() == k and used.count(i) == 0) continue;
    if (tried >= 2) continue;
    tried++;
    pre = 10 * pre, pre += i;
    bool curin = used.count(i);
    if (!curin) used.insert(i);
    int res = go(rem - 1, (match and i == digs[p] - '0'));
    if (!curin) used.erase(i);
    pre /= 10;
    if (res != -1) {
      return res;
    }
  }
  return -1;
}
void solve() {
  int n;
  cin >> n >> k;
  used.clear();
  pre = 0;
  digs = to_string(n);
  int res = go(digs.length(), true);
  cout << res << endl;
}
int main() {
  int t;
  cin >> t;
  while (t) {
    solve();
    t--;
  }
}

```