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: Accepted)

n, k = map(int, input().split())
A = list(map(int, input().split()))

Before = []
After = []

for i in range(n):
    now = A[i]
    before = 0
    after = 0
    for j in range(i):
        if now > A[j]:
            before += 1
    for l in range(i+1,n):
        if now > A[l]:
            after += 1
    Before.append(before)
    After.append(after)

ans = 0
mod = 10**9+7
for i in range(n):
    if k%2 == 0:
        multi_one = int(k/2) * (k-1) % mod
        multi_two = int(k/2) * (k+1) % mod 
    else:
        multi_one = int((k-1)/2) * k % mod
        multi_two = int((k+1)/2) * k % mod
    plus = multi_one*Before[i]%mod + multi_two*After[i]%mod
    ans = (ans + plus) % mod

print(ans)