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 = map(int, input().split())
A = list(map(int, input().split()))
minA = min(A)

count1 = 0
for i in range(N):
    for j in range(i + 1, N):
        if A[i] == minA or A[j] == minA:
            continue
        if A[i] > A[j]:
            count1 += 1

A2 = A + A[::]
# count2 = 0
# for i in range(N * 2):
#     for j in range(i + 1, N * 2):
#         if A2[i] > A2[j]:
#             count2 += 1

count2 = count1
for i in range(N):
    for j in range(i + 1, N * 2):
        # if A[i] == minA or A[j] == minA:
        #     continue
        if A2[i] > A2[j]:
            count2 += 1

A3 = A + A[::] + A[::]
# count3 = 0
# for i in range(N * 3):
#     for j in range(i + 1, N * 3):
#         if A3[i] > A3[j]:
#             count3 += 1

count3 = count2
for i in range(N):
    for j in range(i + 1, N * 3):
        if A3[i] > A3[j]:
            count3 += 1

diff1 = count2 - count1
diff2 = count3 - count2
diffx = diff2 - diff1

bn_last = diff1 + diffx * (K - 2)
bn_sum = ((diff1 + bn_last) * (K - 1)) // 2
an_sum = count1 + bn_sum

if K == 1:
    print(count1)
elif K == 2:
    print(count2)
else:
    print(an_sum % ((10 ** 9) + 7))