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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.InputStream;
import java.io.PrintStream;
import java.util.*;

public class Main {

  public static void main(String[] args) {
    solve(System.in, System.out);
  }

  private static final long MOD = 1000000007L;

  static void solve(InputStream is, PrintStream os) {
    Scanner sc = new Scanner(is);

    /* read */
    int n = sc.nextInt();
    long k = sc.nextLong();

    int[] a = new int[n];
    for (int i = 0; i < n; i++) {
      a[i] = sc.nextInt();
    }

    long result = 0;
    for (int i = 0; i < n; i++) {
      long l = 0;
      long r = 0;

      for (int j = 0; j < i; j++) {
        if (a[i] > a[j]) {
          l++;
//          result += (k - 1 ) * k / 2;
//          result %= MOD;
        }
      }
      result += ((l * (k - 1 )) % MOD * k / 2) % MOD;
      result %= MOD;

      for (int j = i + 1; j < n; j++) {
        if (a[i] > a[j]) {
          r++;
//          result += k * (k + 1) / 2;
//          result %= MOD;
        }
      }
      result += ((r * k) % MOD * (k + 1)) / 2 % MOD;
      result %= MOD;

//      result += calc(r, k, r + l);
//      result %= MOD;
    }

    os.println(result);
  }

  private static long calc(long a, long n, long d) {
    long result = 0;
    result = (result + 2 * a) % MOD;
    result = (result + (n - 1) * d) % MOD;
    result = (result * n) % MOD;
    result = (result / 2) % MOD;
    return result;
//    return ((n * (2 * a + (n - 1) * d)) / 2) % MOD;
  }
}