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)

import sys
input = sys.stdin.readline

N,K=map(int,input().split())
A=list(map(int,input().split()))
mod=10**9+7


B0=[]

for i in range(N):
    ANS=0
    k=A[i]
    for j in range(i+1,N):
        if A[j]<k:
            ANS+=1

    B0.append(ANS)

count=[0]*(max(A)+1)

for a in A:
    count[a]+=1

from itertools import accumulate
S=list(accumulate(count))

ANS=0

for i in range(N):
    t=S[A[i]-1]
    ANS=(ANS+B0[i]*K+t*K*(K-1)//2)%mod

print(ANS)