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<algorithm>
#include<math.h>
#include<queue>
#include<vector>
#include<climits>
#include<map>
#include<string>
#include<functional>
#include<iomanip>
#include<deque>
#include<random>

using namespace std;
typedef long long ll;
typedef double lldo;
#define mp make_pair
#define pub push_back
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define sz(x) int(x.size())
ll gcd(ll a, ll b) { if (a % b == 0) { return b; } else  return gcd(b, a % b); }
ll lcm(ll a, ll b) { if (a == 0) { return b; }return a / gcd(a, b) * b; }
template<class T>ll LBI(vector<T>& ar, T in) { return lower_bound(ar.begin(), ar.end(), in) - ar.begin(); }
template<class T>ll UBI(vector<T>& ar, T in) { return upper_bound(ar.begin(), ar.end(), in) - ar.begin(); }

ll N, K, A[4000] = {}, x1 = 0, x2 = 0;

int main() {
	cin >> N >> K;
	rep(i, N) {
		cin >> A[i];
		A[i + N] = A[i];
	}
	rep(i, N) {
		for (int j = i + 1; j < N * 2; j++) {
			if (A[i] > A[j]) {
				if (j < N) {
					x1++;
				}
				else {
					x2++;
				}
			}
		}
	}
	cout << ((x1* K) % 1000000007 + (x2 * ((K * (K - 1) / 2) % 1000000007)) % 1000000007) % 1000000007 << endl;
	return 0;
}