Polycarp loves geometric progressions very much. Since he was only three years old, he loves only the progressions of length three. He also has a favorite integer k and a sequence a, consisting of n integers.

He wants to know how many subsequences of length three can be selected from a, so that they form a geometric progression with common ratio k.

A subsequence of length three is a combination of three such indexes i1, i2, i3, that 1 ≤ i1 < i2 < i3 ≤ n. That is, a subsequence of length three are such groups of three elements that are not necessarily consecutive in the sequence, but their indexes are strictly increasing.

A geometric progression with common ratio k is a sequence of numbers of the form b·k0, b·k1, ..., b·kr - 1.

Polycarp is only three years old, so he can not calculate this number himself. Help him to do it.

Input

The first line of the input contains two integers, n and k (1 ≤ n, k ≤ 2·105), showing how many numbers Polycarp's sequence has and his favorite number.

The second line contains n integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109) — elements of the sequence.

Output

Output a single number — the number of ways to choose a subsequence of length three, such that it forms a geometric progression with a common ratio k.

Examples

Input

5 2
1 1 2 2 4


Output

4

Input

3 1
1 1 1


Output

1

Input

10 3
1 2 6 2 3 6 9 18 3 9


Output

6

Note

In the first sample test the answer is four, as any of the two 1s can be chosen as the first element, the second element can be any of the 2s, and the third element of the subsequence must be equal to 4.
#include <bits/stdc++.h>
using namespace std;
long long delbit(long long a, int k) { return a & (~(1ll << k)); }
int delbit(int a, int k) { return a & (~(1 << k)); }
bool getbit(long long a, int k) { return 1 & (a >> k); }
bool getbit(int a, int k) { return 1 & (a >> k); }
bool getbit(unsigned long long a, int k) { return 1 & (a >> k); }
long long setbit(long long a, long long k) { return a |= (1 << k); }
int setbit(int a, int k) { return a |= (1 << k); }
inline long long mulmod(long long first, long long n, long long _mod) {
  long long res = 0;
  while (n) {
    if (n & 1) res = (res + first) % _mod;
    first = (first + first) % _mod;
    n >>= 1;
  }
  return res;
}
inline long long powmod(long long first, long long n, long long _mod) {
  long long res = 1;
  while (n) {
    if (n & 1) res = (res * first) % _mod;
    first = (first * first) % _mod;
    n >>= 1;
  }
  return res;
}
inline long long gcd(long long a, long long b) {
  long long t;
  while (b) {
    a = a % b;
    t = a;
    a = b;
    b = t;
  }
  return a;
}
inline int gcd(int a, int b) {
  int t;
  while (b) {
    a = a % b;
    t = a;
    a = b;
    b = t;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long gcd(long long a, long long b, long long c) {
  return gcd(gcd(a, b), c);
}
inline int gcd(int a, int b, int c) { return gcd(gcd(a, b), c); }
template <typename A, typename B>
ostream &operator<<(ostream &stream, const pair<A, B> &p) {
  stream << "{" << p.first << "," << p.second << "}";
  return stream;
}
template <typename A>
ostream &operator<<(ostream &stream, const vector<A> &v) {
  stream << "[";
  for (auto itr = v.begin(); itr != v.end(); itr++) stream << *itr << " ";
  stream << "]";
  return stream;
}
template <typename A, typename B>
ostream &operator<<(ostream &stream, const map<A, B> &v) {
  stream << "[";
  for (auto itr = v.begin(); itr != v.end(); itr++) stream << *itr << " ";
  stream << "]";
  return stream;
}
template <typename A>
ostream &operator<<(ostream &stream, const set<A> &v) {
  stream << "[";
  for (auto itr = v.begin(); itr != v.end(); itr++) stream << *itr << " ";
  stream << "]";
  return stream;
}
template <typename A>
ostream &operator<<(ostream &stream, const priority_queue<A> &v) {
  priority_queue<A> q = v;
  stream << "[";
  while (!q.empty()) {
    stream << q.top() << " ";
    q.pop();
  }
  stream << "]";
  return stream;
}
template <typename A>
ostream &operator<<(ostream &stream, const queue<A> &v) {
  queue<A> q = v;
  stream << "[";
  while (!q.empty()) {
    stream << q.front() << " ";
    q.pop();
  }
  stream << "]";
  return stream;
}
template <typename T, typename T2>
void printarr(T a[], T2 sz, T2 beg = 0) {
  for (T2 i = beg; i < sz; i++) cout << a[i] << " ";
  cout << endl;
}
int a[201001];
long long c[201001];
map<int, long long> cnt, c2;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  for (int(i) = (0); (i) < (n); (i)++) {
    cin >> a[i];
    if (a[i] % k == 0) c[i] += cnt[a[i] / k];
    cnt[a[i]]++;
  }
  long long ans = 0;
  for (int(i) = (0); (i) < (n); (i)++) {
    if (a[i] % k == 0) ans += c2[a[i] / k];
    c2[a[i]] += c[i];
  }
  cout << ans << endl;
  return 0;
}
