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)

MOD = 1000000007
N,K = map(int,input().split())
A = list(map(int,input().split()))
# import numpy as np
# A = np.random.rand(2000)
def bubble_sort(A):
    cnt = 0
    for i in range(len(A)):
        for j in range(len(A)-1, i, -1):
            if A[j] < A[j-1]:
                A[j], A[j-1] = A[j-1], A[j]
                cnt += 1
    return cnt

def sigma(k):
    #return int((1/2)*(k%MOD)*((k+1)%MOD))
    return k*(k+1)//2


# B = []
# for k in range(K):
#     B.extend(A)
# print (bubble_sort(B))


C = []
for k in range(2):
    C.extend(A)
cnt = bubble_sort(A)
#print (cnt)
diff = bubble_sort(C) - 2 * cnt
#print (diff)
a = (K*cnt)
b = (diff*sigma(K-1))
print ((a+b)%MOD)