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: Wrong Answer)

#include <stdio.h>

const int MOD = 1000000007;

int main(void)
{
  int N;
  long long int K;
  scanf("%d %lld", &N, &K);
  int A[N];
  for (int i = 0; i < N; i++)
    scanf("%d", &A[i]);
  int n1 = 0;
  int n2 = 0;
  for (int i = 0; i < N; i++) {
      int Ai = A[i];
      for (int j = 0; j < i; j++)
          if(Ai > A[j])
              n1++;
      for (int j = i+1; j < N; j++)
          if(Ai > A[j])
              n2++;
  }
  printf("K = %ld\nn1 = %d\nn2 = %d\n", K, n1, n2);
  long long int num = K*(K-1)/2%MOD*n1+K*(K+1)/2%MOD*n2;
  printf("%lld\n", num%MOD);
  return 0;
}