codeforces567C. Geometric Progresmit(DP)

C. Geometric Progression

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 21 1 2 2 4

output

4

input

3 11 1 1

output

1

input

10 31 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.

题意：

给你n个数字，选出三个组成公比为k的序列，问你一共有多少组？

思路：

dp[i][j]表示以j结尾并且在三个中位置为i的序列的个数，状态转移方程见代码。

代码：

#include <bits/stdc++.h>using namespace std;typedef long long LL;map<int, LL> dp[4];int n, k, v[200000+100];int main(){    cin >> n >> k;    for (int i = 0; i < n; i++)        scanf("%d", &v[i]);    LL ans = 0;    for (int j = 0; j < n; j++)    {        int x = v[j];//枚举末尾数字        if (x % k == 0)        {            ans += dp[2][x / k];            dp[2][x] += dp[1][x / k];        }        dp[1][x] += 1;    }    cout << ans << endl;    return 0;}