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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <math.h>

int main(void) {

  long mod = pow(10,9)+7;
  long n,k;
  scanf("%ld %ld", &n, &k);
  long a[n];
  for (long i = 0; i < n; i++) {
    scanf("%ld", &a[i]);
  }
  long left[n],right[n];
  for (long i = 0; i < n; i++) {
    left[i] = 0;
    for (long j = 0; j < i; j++) {
      if (a[j] < a[i]) {
        left[i]++;
      }
    }
    right[i] = 0;
    for (long j = i+1; j < n; j++) {
      if (a[j] < a[i]) {
        right[i]++;
      }
    }
  }
  long sum = 0;
  long plus;
  for (long i = 0; i < n; i++) {
    plus = (k-1)*k/2;
    plus %= mod;
    plus *= left[i]+right[i];
    plus %= mod;
    plus += right[i]*k;
    plus %= mod;
    sum += plus;
    sum %= mod;
  }
  printf("%ld\n", sum);

  return 0;
}