#include <bits/stdc++.h>
using namespace std;
const long long MAX = 900000;
const long long MOD = 1000000007;
const long long OO = 0x3f3f3f3f;
const double EPS = 1e-9;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long sum(long long l, long long r) { return (l + r) * (r - l + 1) / 2; }
long long cnt[MAX];
void solve() {
  long long n;
  char c;
  string s;
  cin >> n >> c >> s;
  for (long long i = 0; i <= n; i++) cnt[i] = 0;
  for (long long d = 1; d <= n; d++)
    for (long long i = d; i <= n; i += d)
      if (s[i - 1] != c) cnt[d]++;
  if (count(s.begin(), s.end(), c) == n) {
    cout << "0\n";
    return;
  }
  vector<long long> ans;
  for (long long i = 1; i <= n and ans.empty(); i++)
    if (cnt[i] == 0) ans.push_back(i);
  if (ans.empty()) {
    ans.push_back(n);
    ans.push_back(n - 1);
  }
  cout << ans.size() << '\n';
  for (long long d : ans) cout << d << ' ';
  cout << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t = 1;
  cin >> t;
  for (long long it = 1; it <= t; ++it) {
    solve();
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

Theofanis has a string s_1 s_2 ... s_n and a character c. He wants to make all characters of the string equal to c using the minimum number of operations.

In one operation he can choose a number x (1 ≤ x ≤ n) and for every position i, where i is not divisible by x, replace s_i with c. 

Find the minimum number of operations required to make all the characters equal to c and the x-s that he should use in his operations.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.

The first line of each test case contains the integer n (3 ≤ n ≤ 3 ⋅ 10^5) and a lowercase Latin letter c — the length of the string s and the character the resulting string should consist of.

The second line of each test case contains a string s of lowercase Latin letters — the initial string.

It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.

Output

For each test case, firstly print one integer m — the minimum number of operations required to make all the characters equal to c.

Next, print m integers x_1, x_2, ..., x_m (1 ≤ x_j ≤ n) — the x-s that should be used in the order they are given.

It can be proved that under given constraints, an answer always exists. If there are multiple answers, print any.

Example

Input


3
4 a
aaaa
4 a
baaa
4 b
bzyx


Output


0
1
2
2 
2 3

Note

Let's describe what happens in the third test case: 

  1. x_1 = 2: we choose all positions that are not divisible by 2 and replace them, i. e. bzyx → bzbx; 
  2. x_2 = 3: we choose all positions that are not divisible by 3 and replace them, i. e. bzbx → bbbb. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long MAX = 900000;
const long long MOD = 1000000007;
const long long OO = 0x3f3f3f3f;
const double EPS = 1e-9;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long sum(long long l, long long r) { return (l + r) * (r - l + 1) / 2; }
long long cnt[MAX];
void solve() {
  long long n;
  char c;
  string s;
  cin >> n >> c >> s;
  for (long long i = 0; i <= n; i++) cnt[i] = 0;
  for (long long d = 1; d <= n; d++)
    for (long long i = d; i <= n; i += d)
      if (s[i - 1] != c) cnt[d]++;
  if (count(s.begin(), s.end(), c) == n) {
    cout << "0\n";
    return;
  }
  vector<long long> ans;
  for (long long i = 1; i <= n and ans.empty(); i++)
    if (cnt[i] == 0) ans.push_back(i);
  if (ans.empty()) {
    ans.push_back(n);
    ans.push_back(n - 1);
  }
  cout << ans.size() << '\n';
  for (long long d : ans) cout << d << ' ';
  cout << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t = 1;
  cin >> t;
  for (long long it = 1; it <= t; ++it) {
    solve();
  }
  return 0;
}

```