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: Wrong Answer)

n,k = map(int,input().split())
a = [int(i) for i in input().split()]

mod = 10**9+7

g = 0
for i in range(n):
    for j in range(n):
        if i == j:
            continue
        elif i >= j:
            continue
        if a[i] > a[j]:
            #print(i,j)
            g += 1

if k == 1:
    print(g)
    exit()
    
a2 = []

for i in range(2*n):
    a2.append(a[i%n])
    
f = 0

for i in range(n):
    for j in range(2*n):
        if i == j:
            continue
        elif i >= j:
            continue
        if a2[i] > a2[j]:
            #print(i,j)
            f += 1

t = int((f-g)%mod)
k = int(k%mod)

#print(t)
#exit()
#print(int((k*g%mod + t*(0.5*((k-1)*k))%mod)%mod))
#print(int(( (k%mod*g%mod)%mod + (t*(0.5*((k-1)%mod*k%mod)) ) %mod)%mod))
#print(int((k*g%mod + t*(0.5*((k-1)%mod*k%mod))%mod)%mod))
#print(int((k*g%mod + t*(0.5*((k-1)%mod*k%mod))%mod)%mod))
ans1 = int((k*g%mod)%mod)
#ans2 = t*(0.5* (k-1)%mod * k%mod ) %mod
#ans2 = t*(0.5*((k-1)%mod*k%mod))%mod
ans2 = int(t*(((k-1)%mod*k%mod)%mod)//2%mod)
print( int( (ans1%mod + ans2%mod)%mod ))
#print(int(((k*g) + (t*(0.5*((k-1)*k))))%mod))
exit()