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)

#coding:utf-8
import collections

N, K = list(map(int, input().split()))
A = list(map(int, input().split()))

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

inner_count = count_inversion(A)

A = sorted(A)

duplicate = {k:1 for k,v in enumerate(A)}

A_index = 0
duplicate_index = 0

while A_index < len(A)- 1:
    if A[A_index+1] == A[A_index]:
        duplicate[A_index + 1] -= 1
        duplicate[duplicate_index] += 1
    else:
        duplicate_index = A_index + 1
    A_index += 1
        
count = 0
for k, v in duplicate.items():
    count += (v * k * K * (K-1) )//2

count += inner_count * K

print(count % (10**9 + 7))