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>
using namespace std;
#pragma warning (disable: 4996)

long long mod = 1000000007;
long long N, K, A[100009], B1, B2;

int main() {
	cin >> N >> K;
	for (int i = 0; i < N; i++) cin >> A[i];
	for (int i = 0; i < N; i++) {
		for (int j = i + 1; j < N; j++) {
			if (A[i] > A[j]) B1++;
		}
	}
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			if (A[i] > A[j]) B2++;
		}
	}
	long long C1 = K;
	long long C2 = K * (K - 1) / 2LL; C2 %= mod;
	cout << (B1*C1 + B2 * C2) % mod << endl;
	return 0;
}