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 <bits/stdc++.h>

using namespace std;

const int MOD = 1e9 + 7;

int n, k;
int a[2111];

int main() {
	cin >> n >> k;
	for (int i = 1; i <= n; ++i) {
		cin >> a[i];
	}
	int x = 0, y = 0;
	for (int i = 1; i <= n; ++i) for (int j = i + 1; j <= n; ++j) x += a[i] > a[j];
	x = (long long) x * k % MOD;
	for (int i = 1; i <= n; ++i) for (int j = 1; j <= n; ++j) if (i != j && a[i] > a[j]) y++;
	y = (long long) y * ((long long) k * (k - 1) / 2 % MOD) % MOD;
	cout << ((long long) x + y) % MOD << "\n";
	return 0;
}