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 = [int(i) for i in input().split()]
ans = 0
count = [[0,0] for i in range(N)]
for i in range(N):
  for j in range(i):
    if A[i] > A[j]:
      count[i][0] += 1
  for k in range(i+1,N):
    if A[i] > A[k]:
      count[i][1] += 1
  if K%2 == 0:
    ans += count[i][1]*(K*K//2+K//2) + count[i][0]*((K-1)*((K-1)//2+1))
  else:
    ans += count[i][1]*(K*(K//2+1)) + count[i][0]*((K-1)*((K-1)//2)+(K-1)//2)
print(ans%(10**9+7))