#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
long long st[10][1 << 10];
void solve() {
  int n, kkk;
  scanf("%d%d", &n, &kkk);
  int x = n, len = 0;
  while (x) x /= 10, len++;
  if (kkk == 1) {
    for (int num = 1; num <= 9; num++) {
      long long sum = 0, cur = 1;
      for (int i = 1; i <= len; i++) sum += num * cur, cur *= 10;
      if (sum >= n) {
        printf("%lld\n", sum);
        return;
      }
    }
  } else {
    long long ans = 1e18;
    for (int num1 = 0; num1 <= 9; num1++)
      for (int num2 = num1; num2 <= 9; num2++)
        for (int s = 0; s < (1 << len); s++) {
          long long sum = st[num1][s] + st[num2][(~s) & ((1 << len) - 1)];
          if (sum >= n) ans = min(ans, sum);
        }
    printf("%lld\n", ans);
  }
}
int main() {
  for (int num = 0; num <= 9; num++)
    for (int s = 0; s < (1 << 10); s++) {
      long long sum = 0, cur = 1;
      for (int i = 1; i <= 10; i++)
        sum += ((s & (1 << (i - 1))) ? num : 0) * cur, cur *= 10;
      st[num][s] = sum;
    }
  int T;
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}

### 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>
#pragma GCC optimize("Ofast")
using namespace std;
long long st[10][1 << 10];
void solve() {
  int n, kkk;
  scanf("%d%d", &n, &kkk);
  int x = n, len = 0;
  while (x) x /= 10, len++;
  if (kkk == 1) {
    for (int num = 1; num <= 9; num++) {
      long long sum = 0, cur = 1;
      for (int i = 1; i <= len; i++) sum += num * cur, cur *= 10;
      if (sum >= n) {
        printf("%lld\n", sum);
        return;
      }
    }
  } else {
    long long ans = 1e18;
    for (int num1 = 0; num1 <= 9; num1++)
      for (int num2 = num1; num2 <= 9; num2++)
        for (int s = 0; s < (1 << len); s++) {
          long long sum = st[num1][s] + st[num2][(~s) & ((1 << len) - 1)];
          if (sum >= n) ans = min(ans, sum);
        }
    printf("%lld\n", ans);
  }
}
int main() {
  for (int num = 0; num <= 9; num++)
    for (int s = 0; s < (1 << 10); s++) {
      long long sum = 0, cur = 1;
      for (int i = 1; i <= 10; i++)
        sum += ((s & (1 << (i - 1))) ? num : 0) * cur, cur *= 10;
      st[num][s] = sum;
    }
  int T;
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}

```