#include <bits/stdc++.h>
using namespace std;
long long a, b;
set<long long> s, y;
void func(int l, long long num) {
  if (l <= 0) {
    s.insert(num);
    return;
  }
  func(l - 1, num * 10 + a);
  func(l - 1, num * 10 + b);
}
void fuck(int l, long long num) {
  if (l <= 0) {
    y.insert(num);
    return;
  }
  fuck(l - 1, num * 10 + a);
}
int main() {
  for (int i = 1; i <= 9; i++) s.insert(i);
  for (int i = 2; i <= 11; i++) {
    for (int j = 0; j <= 9; j++) {
      for (int k = 0; k <= 9; k++) {
        a = j, b = k;
        func(i, 0);
      }
    }
  }
  for (int i = 1; i <= 11; i++)
    for (int j = 0; j <= 9; j++) {
      a = j;
      fuck(i, 0);
    }
  int t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    if (k == 2)
      cout << *(s.lower_bound(n)) << endl;
    else
      cout << *(y.lower_bound(n)) << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write 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 a, b;
set<long long> s, y;
void func(int l, long long num) {
  if (l <= 0) {
    s.insert(num);
    return;
  }
  func(l - 1, num * 10 + a);
  func(l - 1, num * 10 + b);
}
void fuck(int l, long long num) {
  if (l <= 0) {
    y.insert(num);
    return;
  }
  fuck(l - 1, num * 10 + a);
}
int main() {
  for (int i = 1; i <= 9; i++) s.insert(i);
  for (int i = 2; i <= 11; i++) {
    for (int j = 0; j <= 9; j++) {
      for (int k = 0; k <= 9; k++) {
        a = j, b = k;
        func(i, 0);
      }
    }
  }
  for (int i = 1; i <= 11; i++)
    for (int j = 0; j <= 9; j++) {
      a = j;
      fuck(i, 0);
    }
  int t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    if (k == 2)
      cout << *(s.lower_bound(n)) << endl;
    else
      cout << *(y.lower_bound(n)) << endl;
  }
  return 0;
}

```