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 Scala (2.11.7) (judge verdict: Accepted)

object Main {

  def main(args: Array[String]): Unit = {
    val sc = new java.util.Scanner(System.in)

    val N = sc.nextInt()
    val K = sc.nextLong()
    val A = Array.fill(N)(sc.nextInt())

    // 左の数の方が大きい数になっている個数
    // 全部見ていくと間に合わない
    // 一旦無視すると,
    // ある数 aiについて
    // aiより大きい数の数を AiC とする
    // aiより左にある大きい数の数を AiCL とする
    // aiより右にある大きい数の数を ACiR とする, ACiR = AiC - AiCL
    // ここで K = 1なら, AiCLであり
    // K = 2なら, AiCL + AiCL + AiCR = AiC + AiCL であり,
    // K >= 3なら, (AiC * (AiC + 1) / 2) - AiC
    // つまり AiCR は必要ない

    val Inf = 1000000000 + 7

    val AiC = new Array[Int](N)
    val AiCL = new Array[Int](N)

    var i = 1
    while (i < N) {
      var j = 0
      while (j < i) {
        if (A(j) > A(i))
          AiCL(i) += 1
        j += 1
      }
      i += 1
    }

    i = 0
    while (i < N) {
      var j = 0
      while (j < N) {
        if (i != j && A(j) > A(i))
          AiC(i) += 1
        j += 1
      }
      i += 1
    }

    var r = BigInt(0)
    i = 0
    while (i < N) {
      r += BigInt(K * (K - 1L) / 2L) * AiC(i).toLong + AiCL(i).toLong * K
      i += 1
    }

    println(r % Inf.toLong)
  }

}