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)

MOD = 1**9+7

def resolve():
  ans = 0  
  n, k = map(int, input().split())
  a = list(map(int, input().split()))
  c = [0] * n  # 自分より小さい値が全体に何個あるか
  for i in range(n):
    for j in range(n):
      if a[i] > a[j]:
        c[i] += 1
        if i < j:    # A 内に存在する転倒数
          ans += k
          ans %= MOD

  s = (k * (k - 1)) // 2
  s %= MOD

  for v in c:
      ans += (v * s) % MOD
      ans %= MOD
  print(int(ans % MOD))
  return  

if __name__ == "__main__":
  resolve()