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)


def main2():
    len_a, kurikaesi = map(int, input().split(" "))
    A = list(map(int, input().split(" ")))

    a_nomi_tenti = 0
    iko_tonari_tenti = 0
    last = 0
    last_tonari = 0
    for index, a_base in enumerate(A):
        for ind2, more_right in enumerate(A[index+1:]):
            if a_base > more_right:
                a_nomi_tenti += 1
        for ind, a_youso in enumerate(A):
            if a_base > a_youso:
                iko_tonari_tenti += 1
    m = (kurikaesi-1)
    kosu = int((1+m) * m / 2) % (10**9+7)
    result = (a_nomi_tenti * kurikaesi) % (10**9+7) + (iko_tonari_tenti * kosu) % (10**9+7)
    print(result % (10**9+7))


main2()