You are given array a with n elements and the number m. Consider some subsequence of a and the value of least common multiple (LCM) of its elements. Denote LCM as l. Find any longest subsequence of a with the value l ≤ m.

A subsequence of a is an array we can get by erasing some elements of a. It is allowed to erase zero or all elements.

The LCM of an empty array equals 1.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 106) — the size of the array a and the parameter from the problem statement.

The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of a.

Output

In the first line print two integers l and kmax (1 ≤ l ≤ m, 0 ≤ kmax ≤ n) — the value of LCM and the number of elements in optimal subsequence.

In the second line print kmax integers — the positions of the elements from the optimal subsequence in the ascending order.

Note that you can find and print any subsequence with the maximum length.

Examples

Input

7 8
6 2 9 2 7 2 3


Output

6 5
1 2 4 6 7


Input

6 4
2 2 2 3 3 3


Output

2 3
1 2 3
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T SQR(T x) {
  return x * x;
}
template <typename T>
inline T MIN(T x, T y) {
  return (x < y) ? x : y;
}
template <typename T>
inline T MAX(T x, T y) {
  return (x > y) ? x : y;
}
template <typename T>
inline T ABS(T x) {
  return (x > 0) ? x : -x;
}
template <typename T>
inline void UPDATE_MIN(T &x, T y) {
  if (y < x) {
    x = y;
  }
}
template <typename T>
inline void UPDATE_MAX(T &x, T y) {
  if (x < y) {
    x = y;
  }
}
template <typename T>
inline int ARGMAX(T cont) {
  return max_element(cont.begin(), cont.end()) - cont.begin();
}
template <typename T>
inline int ARGMIN(T cont) {
  return min_element(cont.begin(), cont.end()) - cont.begin();
}
vector<string> split(const string &s, char c) {
  vector<string> v;
  stringstream ss(s);
  string x;
  while (getline(ss, x, c)) v.emplace_back(x);
  return move(v);
}
template <typename T, typename... Args>
inline string arrStr(T arr, int n) {
  stringstream s;
  s << "[";
  for (__typeof(n - 1) i = (0), __tmpvar__end87 = (n - 1); i < __tmpvar__end87;
       i++)
    s << arr[i] << ",";
  s << arr[n - 1] << "]";
  return s.str();
}
template <class T>
inline string TOSTR(const T &x) {
  stringstream ss;
  ss << x;
  return ss.str();
}
inline int gcd(int a, int b) { return a ? gcd(b % a, a) : b; }
inline long long gcd(long long a, long long b) { return a ? gcd(b % a, a) : b; }
inline long long powmod(long long a, long long p, long long m) {
  long long r = 1;
  while (p) {
    if (p & 1) r = r * a % m;
    p >>= 1;
    a = a * a % m;
  }
  return r;
}
struct pairhash {
  template <typename T, typename U>
  std::size_t operator()(const std::pair<T, U> &x) const {
    return std::hash<T>()(x.first) ^ std::hash<U>()(x.second);
  }
};
template <typename K, typename V>
V GetWithDef(const std::unordered_map<K, V> &m, const K &key, const V &defval) {
  auto it = m.find(key);
  return (it == m.end()) ? defval : it->second;
}
template <typename K, typename V>
void SetDef(std::unordered_map<K, V> &m, const K &key, const V &defval) {
  auto it = m.find(key);
  if (it == m.end()) m[key] = defval;
}
const int MOD = 1000 * 1000 * 1000 + 7;
const double PI = 3.1415926535897932384626433832795l;
inline void addto(int &a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
inline int add(int a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
  return a;
}
inline void subto(int &a, int b) {
  a -= b;
  if (a < 0) a += MOD;
  if (a >= MOD) a -= MOD;
}
inline int sub(int a, int b) {
  a -= b;
  if (a < 0) a += MOD;
  if (a >= MOD) a -= MOD;
  return a;
}
inline void multo(int &a, int b) { a = (long long)a * b % MOD; }
inline int mul(int a, int b) { return (long long)a * b % MOD; }
inline int mulmod(int a, int b, int mod) { return (long long)a * b % mod; }
inline int powmod(int a, int e, int mod) {
  int x;
  for (x = 1; e > 0; e >>= 1) {
    if (e & 1) x = mulmod(x, a, mod);
    a = mulmod(a, a, mod);
  }
  return x;
}
inline int invmod(int a, int mod) { return powmod(a, mod - 2, mod); }
inline long long invmodLL(long long p) {
  long long q = p;
  for (long long a = p * p; a != 1; a *= a) q *= a;
  return q;
}
int N, M, K, L, E;
int arr[1000123] = {};
int amax = -1;
int cnt[1000123] = {};
int bestlen[1000123] = {};
int main() {
  ios_base::sync_with_stdio(0);
  scanf("%d%d", &N, &M);
  for (__typeof(N) i = (0), __tmpvar__end220 = (N); i < __tmpvar__end220; i++) {
    scanf("%d", &arr[i]);
    if (arr[i] > M) continue;
    cnt[arr[i]]++;
    UPDATE_MAX(amax, arr[i]);
  }
  for (__typeof(amax + 1) k = (2), __tmpvar__end226 = (amax + 1);
       k < __tmpvar__end226; k++) {
    for (int cur = k; cur <= M; cur += k) {
      bestlen[cur] += cnt[k];
    }
  }
  int ans_lcm = 1;
  int ans_len = 0;
  for (__typeof(M + 1) lcm = (1), __tmpvar__end233 = (M + 1);
       lcm < __tmpvar__end233; lcm++) {
    if (bestlen[lcm] > ans_len) {
      ans_len = bestlen[lcm];
      ans_lcm = lcm;
    }
  }
  vector<int> ans;
  for (__typeof(N) i = (0), __tmpvar__end240 = (N); i < __tmpvar__end240; i++) {
    if (ans_lcm % arr[i] == 0) ans.push_back(i + 1);
  }
  sort((ans).begin(), (ans).end());
  printf("%d %d\n", ans_lcm, ans_len + cnt[1]);
  for (auto &i : ans) printf("%d ", i);
  puts("");
  return 0;
}
