#include <bits/stdc++.h>
using namespace std;
vector<long long> g[11], f;
void dfs(long long num, int a, int b, int l, int mdl) {
  if (l == 0) {
    g[mdl].push_back(num);
    return;
  }
  dfs(num * 10 + a, a, b, l - 1, mdl);
  dfs(num * 10 + b, a, b, l - 1, mdl);
}
void init() {
  for (int i = 0; i <= 9; i++) {
    long long num = 0;
    for (int j = 1; j <= 10; j++) {
      num = num * 10 + i;
      g[j].push_back(num), f.push_back(num);
    }
  }
  for (int l = 1; l <= 10; l++) {
    for (int i = 0; i <= 9; i++)
      for (int j = i + 1; j <= 9; j++) dfs(0, i, j, l, l);
  }
  for (int l = 1; l <= 10; l++) sort(g[l].begin(), g[l].end());
}
int main() {
  init();
  int T;
  cin >> T;
  while (T--) {
    char s[12];
    long long o = 0;
    int k;
    scanf("%s", s + 1);
    int l = strlen(s + 1);
    for (int j = 1; j <= l; j++) o = o * 10 + (s[j] - '0');
    scanf("%d", &k);
    if (k == 1) {
      long long ans = 100000000000ll;
      for (int j = 0; j < f.size(); j++) {
        if (f[j] >= o) ans = min(ans, f[j]);
      }
      printf("%lld\n", ans);
    } else {
      for (int j = 0; j < g[l].size(); j++)
        if (g[l][j] >= o) {
          printf("%lld\n", g[l][j]);
          break;
        }
    }
  }
  return 0;
}

### Prompt

Generate 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;
vector<long long> g[11], f;
void dfs(long long num, int a, int b, int l, int mdl) {
  if (l == 0) {
    g[mdl].push_back(num);
    return;
  }
  dfs(num * 10 + a, a, b, l - 1, mdl);
  dfs(num * 10 + b, a, b, l - 1, mdl);
}
void init() {
  for (int i = 0; i <= 9; i++) {
    long long num = 0;
    for (int j = 1; j <= 10; j++) {
      num = num * 10 + i;
      g[j].push_back(num), f.push_back(num);
    }
  }
  for (int l = 1; l <= 10; l++) {
    for (int i = 0; i <= 9; i++)
      for (int j = i + 1; j <= 9; j++) dfs(0, i, j, l, l);
  }
  for (int l = 1; l <= 10; l++) sort(g[l].begin(), g[l].end());
}
int main() {
  init();
  int T;
  cin >> T;
  while (T--) {
    char s[12];
    long long o = 0;
    int k;
    scanf("%s", s + 1);
    int l = strlen(s + 1);
    for (int j = 1; j <= l; j++) o = o * 10 + (s[j] - '0');
    scanf("%d", &k);
    if (k == 1) {
      long long ans = 100000000000ll;
      for (int j = 0; j < f.size(); j++) {
        if (f[j] >= o) ans = min(ans, f[j]);
      }
      printf("%lld\n", ans);
    } else {
      for (int j = 0; j < g[l].size(); j++)
        if (g[l][j] >= o) {
          printf("%lld\n", g[l][j]);
          break;
        }
    }
  }
  return 0;
}

```