Problem:
Score : 300 points

Problem Statement

We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}.

Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2.

Find the inversion number of B, modulo 10^9 + 7.

Here the inversion number of B is defined as the number of ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j.

Constraints

All values in input are integers.

1 \leq N \leq 2000

1 \leq K \leq 10^9

1 \leq A_i \leq 2000

Input

Input is given from Standard Input in the following format:

N K
A_0 A_1 ... A_{N - 1}

Output

Print the inversion number of B, modulo 10^9 + 7.

Sample Input 1

2 2
2 1

Sample Output 1

3

In this case, B~=~2,~1,~2,~1. We have:

B_0 > B_1

B_0 > B_3

B_2 > B_3

Thus, the inversion number of B is 3.

Sample Input 2

3 5
1 1 1

Sample Output 2

0

A may contain multiple occurrences of the same number.

Sample Input 3

10 998244353
10 9 8 7 5 6 3 4 2 1

Sample Output 3

185297239

Be sure to print the output modulo 10^9 + 7.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <cmath>

using namespace std;

int mod = 1000000007;

long long f(int a, int k);

int main()
{
	long long n, k;
	cin >> n >> k;
	vector<int> a(n);
	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}

	vector<pair<long long, long long>> b(n);
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < i; j++) {
			if (a[i] > a[j]) {
				b[i].first++;
			}
		}
		for (int j = i + 1; j < n; j++) {
			if (a[i] > a[j]) {
				b[i].second++;
			}
		}
	}

	long long ans = 0;
	for (int i = 0; i < n; i++) {
		ans += f(b[i].first, k - 1) + f(b[i].second, k);
		ans %= mod;
	}
	cout << ans << endl;

	return 0;
}

long long f(int a, int k)
{
	long long calc = (long long)k * (k + 1) / 2;
	calc %= mod;
	calc *= a;
	calc %= mod;

	return calc;
}