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 P = 1e9 + 7;
int main() {
	int n;
	long long k;
	cin >> n >> k;
	vector<int> a(n);
	for(int i = 0; i < n; ++i) {
		cin >> a[i];
	}
	long long ans = 0;
	for(int i = 0; i < n; ++i) {
		long long tmp = 0;
		for(int j = 0; j < i; ++j) {
			if(a[i] < a[j]) {
				++tmp;
			}
		}
		ans = (ans + tmp * k % P) % P;
		tmp = 0;
		for(int j = 0; j < n; ++j) {
			if(a[i] < a[j]) {
				++tmp;
			}
		}
		ans = (ans + (k * (k - 1) / 2) % P * tmp % P) % P;
	}
	cout << ans % P << '\n';
	return 0;
}