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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

fun main(args: Array<String>) {
    val mod = 1000000007L
    
    val (N, K) = readInputLine().split(" ").map { it.toInt() }
    
    val AArr = readInputLine().split(" ").map { it.toInt() }
    
    var ans = 0L
    
    for (i in 0 until N) {
        var befCnt = 0L
        var aftCnt = 0L
        for (j in 0 until N) {
            if (AArr[i] > AArr[j]) {
                if (i > j) {
                    aftCnt++
                } else {
                    befCnt++
                }
            }
        }

        ans += befCnt * K.toLong() * (K + 1).toLong() / 2L
        ans %= mod
        ans += aftCnt * (K - 1).toLong() * K.toLong() / 2L
        ans %= mod
    }

    println(ans)
}

fun readInputLine(): String {
    return readLine()!!
}