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 Python (3.4.3) (judge verdict: Wrong Answer)

import numpy as np
n, k = map(int, input().split())
a = list(map(int, input().split()))
num = [[0, 0] for _ in range(n)]
for i in range(n):
    for j in range(n):
        if a[i] > a[j]:
            num[i][1] += 1
        if (i < j) and (a[i] > a[j]):
            num[i][0] += 1
num_np = np.array(num)
ks = k * (k - 1) // 2
sum_left = sum(num_np[:,0])
sum_right = sum(num_np[:,1])
output = (sum_left * k) % 1000000007 + (sum_right * ks) % 1000000007
print(int(output) % 1000000007)