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>
#include <stdio.h>

using namespace std;
 
#define int long long
 
bool codejam = 0;
 
typedef long long li;
typedef long double ld;
 
void solve();

int32_t main () {
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	cout << fixed << setprecision(20);
	#ifdef LOCAL
		freopen("input.txt", "r", stdin);
	#endif
	int t = 1; 
	// cin >> t;
	int tt = t;
	while (t--) { 
		if (codejam) cout << "Case #" << tt - t << ": ";
		solve();
	}
	return 0;
}
const int N = 2005;
int arr[N];
const int M = 1e9 + 7;
void solve() {
	int n, k;
	cin >> n >> k;
	for (int i = 1; i <= n; i += 1) {
		cin >> arr[i];
	}
	int ans = 0;
	for (int i = 1; i <= n; i += 1) {
		for (int j = i + 1; j <= n; j += 1) {
			if (arr[i] < arr[j]) {
				ans += (k * (k - 1)) / 2;
				ans %= M;
			}
			else if (arr[i] > arr[j]) {
				ans += (k * (k + 1)) / 2;
				ans %= M;	
			}
		}
	}
	cout << ans;
}