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 PyPy3 (2.4.0) (judge verdict: Accepted)

N, K = map(int, input().split())
A = list(map(int, input().split()))
MOD = 10**9+7
nums = A[:]
nums_sorted = sorted(nums)
rank = {}
count = 0
for n in nums_sorted:
    if not n in rank:
        rank[n] = count
        count += 1
    else:
        count += 1
#print(rank)

def count_inversion(sequence):
    count = 0
    ret = [0] * len(sequence)
    for i in range(1,len(sequence),1):
        for j in range(i,len(sequence),1):
            if sequence[i-1] > sequence[j]:
                ret[i-1] += 1
    return ret
rev = count_inversion(nums)

#print(rev)
ans = 0
n = K
for a, d in zip(rev, nums):
    ans += n*(2*a + (n-1)*rank[d])//2 % MOD
print(ans% MOD)