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)

N,K = [int(i) for i in input().split()]
A_list = [int(i) for i in input().split()]

mod = 10**9+7


rightlist = [0 for i in A_list]
alllist = [0 for i in A_list]

for i in range(len(A_list)):
    ct1 = 0
    ct2 = 0
    a = A_list[i]
    right = A_list[i:]
    
    for j in right:
        #print(a,j)
        if a > j:
            ct1+=1
    
    for j in A_list:
        if a > j:
            ct2+=1
    rightlist[i] = ct1
    alllist[i] = ct2


R = int(sum(rightlist))
A  =int(sum(alllist))
K = int(K%mod)
R = int(R%mod)
A = int(A%mod)

ans = int(int(R*K)+A*int(K*(K-1)*2**-1))
print(ans%mod)