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 (1.3.71) (judge verdict: Accepted)

fun main(args: Array<String>){
    val (N,K) = readLine()!!.split(" ").map{ it.toLong()}
    var A = readLine()!!.split(" ").map{it.toInt()}
    var preOK = LongArray(N.toInt()) {0}
    var afterOK = LongArray(N.toInt()) {0}
    var wholeOK = LongArray(N.toInt()) {0}
    val MOD = 1000000007


    for(i in 0 until A.size){
        for(j in 0 until A.size){
            if(i > j && A[i] > A[j]){
                preOK[i]++
            }  else if(i < j && A[i] > A[j]) {
                afterOK[i]++
            }
        }
        wholeOK[i] = preOK[i] + afterOK[i]
    }
    var ans =0L

    for(i in 0 until A.size){
        ans += afterOK[i]*K
        ans %= MOD
        ans += ((K*(K-1)/2L) % MOD)  * wholeOK[i]
        ans %= MOD
    }


    while(ans<0) ans += MOD



    println(ans)
}