#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 1;
const int MOD = 1e9 + 7;
const double EPS = 1e-8;
set<pair<int, string>> S;
void search(string cur, int i, int j) {
  S.insert({cur.size(), cur});
  if (cur.size() >= 10) return;
  if (cur.size() || i != 0) search(cur + char(i + '0'), i, j);
  if (cur.size() != 0 || j != 0) search(cur + char(j + '0'), i, j);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  for (int i = 0; i <= 9; ++i)
    for (int j = i; j <= 9; ++j) search("", i, j);
  int tt;
  cin >> tt;
  while (tt--) {
    string n, temp = "";
    int k;
    cin >> n >> k;
    if (k == 1) {
      for (int i = 1; i <= 9; ++i) {
        for (int j = 0; j < n.size(); ++j) {
          temp += char(i + '0');
        }
        if (temp >= n) {
          cout << temp;
          break;
        }
        temp = "";
      }
    } else {
      cout << S.lower_bound({n.size(), n})->second;
    }
    cout << '\n';
  }
}

### Prompt

In Cpp, your task is to solve 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 = 1e5 + 1;
const int MOD = 1e9 + 7;
const double EPS = 1e-8;
set<pair<int, string>> S;
void search(string cur, int i, int j) {
  S.insert({cur.size(), cur});
  if (cur.size() >= 10) return;
  if (cur.size() || i != 0) search(cur + char(i + '0'), i, j);
  if (cur.size() != 0 || j != 0) search(cur + char(j + '0'), i, j);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  for (int i = 0; i <= 9; ++i)
    for (int j = i; j <= 9; ++j) search("", i, j);
  int tt;
  cin >> tt;
  while (tt--) {
    string n, temp = "";
    int k;
    cin >> n >> k;
    if (k == 1) {
      for (int i = 1; i <= 9; ++i) {
        for (int j = 0; j < n.size(); ++j) {
          temp += char(i + '0');
        }
        if (temp >= n) {
          cout << temp;
          break;
        }
        temp = "";
      }
    } else {
      cout << S.lower_bound({n.size(), n})->second;
    }
    cout << '\n';
  }
}

```