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>
#include <math.h>

long INVERSION_NUMBER(int N, int *Array);
int main(void) {

   int N,K;
   
   scanf("%d%d",&N,&K);
   
   int A[N];
   int B[2*N];
   int i;
   for (i = 0; i < N; i++) {
      scanf("%d", &A[i]);
   }
   
   for (i = 0; i < N; i++) {
      B[i] = A[i];
   }
   
   long a1 = INVERSION_NUMBER(N, B);

   for (i = 0; i < N; i++) {
      B[i] = A[i];
   }
   for (i = 0; i < N; i++) {
      B[i+N] = A[i];
   }
   
   long a2 = INVERSION_NUMBER(2*N, B);
   long mix = a2 - a1 - a1;
   long w = (long)pow(10,9) + 7;
   

   printf("%d\n", ((K*a1%w) + ((K - 1)*(K - 0)%w*mix%w/2))%w);
  
   return 0;
}

long INVERSION_NUMBER(int N, int *Array) {
   
   long i,j;
   int temp;
   long ans = 0;
   long count;
   
   for (i = 0;i < N; i++) {
      count = 0;
      for (j = 0; j < N - 1 - i; j++) {
         if (Array[j] > Array[j+1]) {
            temp       = Array[j];
            Array[j]   = Array[j+1];
            Array[j+1] = temp;
            ans++;
            count = 1;
         }
      }
      if (count == 0) {
         break;
      }
   }
   
   return ans;
   
}