#include <bits/stdc++.h>
using namespace std;
vector<long long> primes, fact, facti;
pair<long long, long long> hashstr(string& s, long long l = 0,
                                   long long r = -1) {
  long long p = 31, q = 53, pt = 1, qt = 1;
  pair<long long, long long> hash = {0, 0};
  if (r == -1) r = s.length() - 1;
  for (int i = l; i <= r; i++) {
    hash.first = (hash.first + (s[i] - 'a') * pt) % 1000000007;
    hash.second = (hash.second + (s[i] - 'a') * qt) % 1000000007;
    pt = (pt * p) % 1000000007;
    qt = (qt * q) % 1000000007;
  }
  return hash;
}
long long binpow(long long a, long long b, long long m = -1) {
  if (m == -1) m = 1000000007;
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long modi(long long a, long long m = -1) {
  if (m == -1) m = 1000000007;
  return binpow(a, 1000000007 - 2);
}
vector<long long> gen_prime(long long n) {
  vector<long long> prime;
  if (n < 2) return prime;
  prime.push_back(2);
  if (!(n & 1)) n--;
  vector<bool> is_prime(n / 2, true);
  for (long long i = 3; i * i <= n; i += 2) {
    if (!is_prime[(i - 3) / 2]) continue;
    for (long long j = i * i; j <= n; j += i + i) {
      is_prime[(j - 3) / 2] = false;
    }
  }
  for (int i = 0; i < is_prime.size(); i++) {
    if (is_prime[i]) prime.push_back(i * 2 + 3);
  }
  return prime;
}
void gen_factorial(long long n, bool invert) {
  fact.push_back(1);
  if (invert) facti.push_back(modi(1));
  long long f;
  for (long long i = 1; i <= n; i++) {
    f = (fact[i - 1] * i) % 1000000007;
    fact.push_back(f);
    if (invert) {
      facti.push_back(modi(f));
    }
  }
}
void vlInput(vector<long long>& a) {
  for (int i = 0; i < a.size(); i++) {
    cin >> a[i];
  }
}
long long factorial(long long n) {
  if (n == 0) return 1ll;
  return (n * factorial(n - 1)) % 1000000007;
}
struct triplet {
  long long next, val, last;
};
struct custom {
  long long x, y, z;
};
class minComp {
 public:
  bool operator()(custom& a, custom& b) const {
    long long ad = abs(a.x - a.y), bd = abs(b.x - b.y);
    if (ad == bd) {
      return a.x > b.x;
    }
    return ad < bd;
  }
};
long long factor_count(long long n) {
  long long cnt = 0;
  while (n % 2 == 0) {
    cnt++;
    n = n / 2;
  }
  for (int i = 0; primes[i] * primes[i] <= n; i++) {
    while (n % primes[i] == 0) {
      cnt++;
      n = n / primes[i];
    }
  }
  if (n > 2) cnt++;
  return cnt;
}
long long lis(vector<long long>& a) {
  vector<long long> dp(a.size() + 1, 1000000007);
  long long mxl = -1, l;
  dp[0] = -1000000007;
  for (long long i = 0; i < a.size(); i++) {
    long long l = upper_bound(dp.begin(), dp.end(), a[i]) - dp.begin();
    mxl = max(mxl, l);
    if (dp[l - 1] < a[i] && a[i] < dp[l]) dp[l] = a[i];
  }
  return mxl;
}
void build_segment(vector<long long>& a, long long l, long long r,
                   vector<long long>& segment, long long pos) {
  if (l == r) {
    segment[pos] = a[l];
    return;
  }
  long long mid = (l + r) / 2;
  build_segment(a, l, mid, segment, pos * 2);
  build_segment(a, mid + 1, r, segment, pos * 2 + 1);
  segment[pos] = max(segment[pos * 2], segment[pos * 2 + 1]);
}
long long operateSegment(vector<long long>& segment, long long l, long long r,
                         long long tl, long long tr, long long pos) {
  if (l > r) return -1;
  if (l == tl && r == tr) return segment[pos];
  long long mid = (l + r) / 2;
  long long ans, t1;
  if (mid >= tl) {
    t1 = operateSegment(segment, l, mid, tl, min(mid, tr), pos * 2);
  }
  if (mid < tr) {
    ans =
        operateSegment(segment, mid + 1, r, max(mid + 1, tl), tr, pos * 2 + 1);
  }
  ans = max(ans, t1);
  return ans;
}
void updateSegment(vector<pair<long long, long long> >& segment, long long l,
                   long long r, long long up, long long uv, long long p) {
  if (l == r && l == up) {
    segment[p].first = uv;
    return;
  }
  long long mid = (l + r) / 2;
  if (up <= mid) {
    updateSegment(segment, l, mid, up, uv, p * 2);
  } else {
    updateSegment(segment, mid + 1, r, up, uv, p * 2 + 1);
  }
  segment[p].first = segment[p * 2].first + segment[p * 2 + 1].first;
  segment[p].second = segment[p * 2].second;
  if (segment[p * 2 + 1].first > 0)
    segment[p].second = segment[p * 2 + 1].second;
}
bool sort_input(pair<long long, long long>& a, pair<long long, long long>& b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first > b.first;
}
bool myComp(char& a, char& b) { return a > b; }
int getUniqueUserIdSum(vector<int> arr) {
  int mx = 6000, n = arr.size();
  vector<int> cnt(mx, 0);
  for (auto i : arr) {
    cnt[i]++;
  }
  int ans = 0;
  int left = 0;
  for (int i = 0; i < mx; i++) {
    if (cnt[i] > 0) {
      cnt[i]--;
      left += cnt[i];
      ans += i;
    } else if (left > 0) {
      ans += i;
      left--;
    }
  }
  return ans;
}
long carParkingRoof(vector<long> a, long k) {
  sort(a.begin(), a.end());
  long INF = 1e9 + 7;
  long n = a.size();
  long ans = INF, sum = 0;
  for (int i = k - 1; i < n; i++) {
    sum = a[i] - a[i - k + 1] + 1;
    ans = min(ans, sum);
  }
  if (ans == INF) ans = 0;
  return ans;
}
int finalInstances(int n, vector<int> a) {
  int watch = 0;
  int mx = 1e8;
  for (auto i : a) {
    if (watch > 0) {
      watch--;
      continue;
    }
    if (i > 60 && n <= mx) {
      n *= 2;
      watch = 10;
    } else if (i < 25 && n > 1) {
      n = (n + 1) / 2;
      watch = 10;
    }
  }
  return n;
}
int exam(vector<int> a) {
  int n = a.size();
  vector<int> p(n + 1, 0);
  for (int i = n - 1; i >= 0; i--) {
    p[i] = p[i + 1];
    if (a[i] == 1)
      p[i]++;
    else
      p[i]--;
  }
  int yr = 0;
  for (int i = 0; i < n; i++) {
    if (yr > p[i]) return i;
    if (a[i] == 1)
      yr++;
    else
      yr--;
  }
  return -1;
}
vector<string> extract(string a) {
  vector<string> ans;
  string w;
  for (auto i : a) {
    if (i == ' ') {
      ans.push_back(w);
      w = "";
    } else {
      w.push_back(i);
    }
  }
  return ans;
}
vector<string> pl(vector<string> l, long long t) {
  map<string, long long> m;
  for (auto i : l) {
    vector<string> v = extract(i);
    m[v[0]]++;
    if (v[0] != v[1]) m[v[1]]++;
  }
  vector<string> ans;
  for (auto i : m) {
    if (i.second >= t) {
      ans.push_back(i.first);
    }
  }
  return ans;
}
void solve() {
  long long n;
  char k;
  string s;
  cin >> n >> k >> s;
  vector<long long> r;
  for (long long i = 0; i < n; i++) {
    if (s[i] == k) {
      r.push_back(i + 1);
    }
  }
  if (r.size() == n) {
    cout << "0\n";
    return;
  }
  vector<bool> iwd(n + 1, true);
  for (int m = r.size() - 1; m >= 0; m--) {
    long long i = r[m];
    iwd[i] = false;
    for (long long j = i + i; j <= n; j += i) {
      if (iwd[j]) {
        iwd[i] = true;
        break;
      }
    }
  }
  long long isOne = -1;
  for (auto i : r) {
    if (!iwd[i]) {
      isOne = i;
      break;
    }
  }
  if (isOne != -1)
    cout << "1\n" << isOne << "\n";
  else
    cout << "2\n" << n - 1 << " " << n << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t, i = 0;
  cin >> t;
  for (i = 1; i <= t; i++) solve();
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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;
vector<long long> primes, fact, facti;
pair<long long, long long> hashstr(string& s, long long l = 0,
                                   long long r = -1) {
  long long p = 31, q = 53, pt = 1, qt = 1;
  pair<long long, long long> hash = {0, 0};
  if (r == -1) r = s.length() - 1;
  for (int i = l; i <= r; i++) {
    hash.first = (hash.first + (s[i] - 'a') * pt) % 1000000007;
    hash.second = (hash.second + (s[i] - 'a') * qt) % 1000000007;
    pt = (pt * p) % 1000000007;
    qt = (qt * q) % 1000000007;
  }
  return hash;
}
long long binpow(long long a, long long b, long long m = -1) {
  if (m == -1) m = 1000000007;
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long modi(long long a, long long m = -1) {
  if (m == -1) m = 1000000007;
  return binpow(a, 1000000007 - 2);
}
vector<long long> gen_prime(long long n) {
  vector<long long> prime;
  if (n < 2) return prime;
  prime.push_back(2);
  if (!(n & 1)) n--;
  vector<bool> is_prime(n / 2, true);
  for (long long i = 3; i * i <= n; i += 2) {
    if (!is_prime[(i - 3) / 2]) continue;
    for (long long j = i * i; j <= n; j += i + i) {
      is_prime[(j - 3) / 2] = false;
    }
  }
  for (int i = 0; i < is_prime.size(); i++) {
    if (is_prime[i]) prime.push_back(i * 2 + 3);
  }
  return prime;
}
void gen_factorial(long long n, bool invert) {
  fact.push_back(1);
  if (invert) facti.push_back(modi(1));
  long long f;
  for (long long i = 1; i <= n; i++) {
    f = (fact[i - 1] * i) % 1000000007;
    fact.push_back(f);
    if (invert) {
      facti.push_back(modi(f));
    }
  }
}
void vlInput(vector<long long>& a) {
  for (int i = 0; i < a.size(); i++) {
    cin >> a[i];
  }
}
long long factorial(long long n) {
  if (n == 0) return 1ll;
  return (n * factorial(n - 1)) % 1000000007;
}
struct triplet {
  long long next, val, last;
};
struct custom {
  long long x, y, z;
};
class minComp {
 public:
  bool operator()(custom& a, custom& b) const {
    long long ad = abs(a.x - a.y), bd = abs(b.x - b.y);
    if (ad == bd) {
      return a.x > b.x;
    }
    return ad < bd;
  }
};
long long factor_count(long long n) {
  long long cnt = 0;
  while (n % 2 == 0) {
    cnt++;
    n = n / 2;
  }
  for (int i = 0; primes[i] * primes[i] <= n; i++) {
    while (n % primes[i] == 0) {
      cnt++;
      n = n / primes[i];
    }
  }
  if (n > 2) cnt++;
  return cnt;
}
long long lis(vector<long long>& a) {
  vector<long long> dp(a.size() + 1, 1000000007);
  long long mxl = -1, l;
  dp[0] = -1000000007;
  for (long long i = 0; i < a.size(); i++) {
    long long l = upper_bound(dp.begin(), dp.end(), a[i]) - dp.begin();
    mxl = max(mxl, l);
    if (dp[l - 1] < a[i] && a[i] < dp[l]) dp[l] = a[i];
  }
  return mxl;
}
void build_segment(vector<long long>& a, long long l, long long r,
                   vector<long long>& segment, long long pos) {
  if (l == r) {
    segment[pos] = a[l];
    return;
  }
  long long mid = (l + r) / 2;
  build_segment(a, l, mid, segment, pos * 2);
  build_segment(a, mid + 1, r, segment, pos * 2 + 1);
  segment[pos] = max(segment[pos * 2], segment[pos * 2 + 1]);
}
long long operateSegment(vector<long long>& segment, long long l, long long r,
                         long long tl, long long tr, long long pos) {
  if (l > r) return -1;
  if (l == tl && r == tr) return segment[pos];
  long long mid = (l + r) / 2;
  long long ans, t1;
  if (mid >= tl) {
    t1 = operateSegment(segment, l, mid, tl, min(mid, tr), pos * 2);
  }
  if (mid < tr) {
    ans =
        operateSegment(segment, mid + 1, r, max(mid + 1, tl), tr, pos * 2 + 1);
  }
  ans = max(ans, t1);
  return ans;
}
void updateSegment(vector<pair<long long, long long> >& segment, long long l,
                   long long r, long long up, long long uv, long long p) {
  if (l == r && l == up) {
    segment[p].first = uv;
    return;
  }
  long long mid = (l + r) / 2;
  if (up <= mid) {
    updateSegment(segment, l, mid, up, uv, p * 2);
  } else {
    updateSegment(segment, mid + 1, r, up, uv, p * 2 + 1);
  }
  segment[p].first = segment[p * 2].first + segment[p * 2 + 1].first;
  segment[p].second = segment[p * 2].second;
  if (segment[p * 2 + 1].first > 0)
    segment[p].second = segment[p * 2 + 1].second;
}
bool sort_input(pair<long long, long long>& a, pair<long long, long long>& b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first > b.first;
}
bool myComp(char& a, char& b) { return a > b; }
int getUniqueUserIdSum(vector<int> arr) {
  int mx = 6000, n = arr.size();
  vector<int> cnt(mx, 0);
  for (auto i : arr) {
    cnt[i]++;
  }
  int ans = 0;
  int left = 0;
  for (int i = 0; i < mx; i++) {
    if (cnt[i] > 0) {
      cnt[i]--;
      left += cnt[i];
      ans += i;
    } else if (left > 0) {
      ans += i;
      left--;
    }
  }
  return ans;
}
long carParkingRoof(vector<long> a, long k) {
  sort(a.begin(), a.end());
  long INF = 1e9 + 7;
  long n = a.size();
  long ans = INF, sum = 0;
  for (int i = k - 1; i < n; i++) {
    sum = a[i] - a[i - k + 1] + 1;
    ans = min(ans, sum);
  }
  if (ans == INF) ans = 0;
  return ans;
}
int finalInstances(int n, vector<int> a) {
  int watch = 0;
  int mx = 1e8;
  for (auto i : a) {
    if (watch > 0) {
      watch--;
      continue;
    }
    if (i > 60 && n <= mx) {
      n *= 2;
      watch = 10;
    } else if (i < 25 && n > 1) {
      n = (n + 1) / 2;
      watch = 10;
    }
  }
  return n;
}
int exam(vector<int> a) {
  int n = a.size();
  vector<int> p(n + 1, 0);
  for (int i = n - 1; i >= 0; i--) {
    p[i] = p[i + 1];
    if (a[i] == 1)
      p[i]++;
    else
      p[i]--;
  }
  int yr = 0;
  for (int i = 0; i < n; i++) {
    if (yr > p[i]) return i;
    if (a[i] == 1)
      yr++;
    else
      yr--;
  }
  return -1;
}
vector<string> extract(string a) {
  vector<string> ans;
  string w;
  for (auto i : a) {
    if (i == ' ') {
      ans.push_back(w);
      w = "";
    } else {
      w.push_back(i);
    }
  }
  return ans;
}
vector<string> pl(vector<string> l, long long t) {
  map<string, long long> m;
  for (auto i : l) {
    vector<string> v = extract(i);
    m[v[0]]++;
    if (v[0] != v[1]) m[v[1]]++;
  }
  vector<string> ans;
  for (auto i : m) {
    if (i.second >= t) {
      ans.push_back(i.first);
    }
  }
  return ans;
}
void solve() {
  long long n;
  char k;
  string s;
  cin >> n >> k >> s;
  vector<long long> r;
  for (long long i = 0; i < n; i++) {
    if (s[i] == k) {
      r.push_back(i + 1);
    }
  }
  if (r.size() == n) {
    cout << "0\n";
    return;
  }
  vector<bool> iwd(n + 1, true);
  for (int m = r.size() - 1; m >= 0; m--) {
    long long i = r[m];
    iwd[i] = false;
    for (long long j = i + i; j <= n; j += i) {
      if (iwd[j]) {
        iwd[i] = true;
        break;
      }
    }
  }
  long long isOne = -1;
  for (auto i : r) {
    if (!iwd[i]) {
      isOne = i;
      break;
    }
  }
  if (isOne != -1)
    cout << "1\n" << isOne << "\n";
  else
    cout << "2\n" << n - 1 << " " << n << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t, i = 0;
  cin >> t;
  for (i = 1; i <= t; i++) solve();
  return 0;
}

```