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>

int main(){
   int i,j,n,a[2000],t[2000],ta[2000];
   long s,kk,kk1,k;
 
   scanf("%d %ld",&n,&k);
   for (i=0;i<n;i++) scanf("%d",&a[i]);
   for (i=0;i<n;i++) {
      t[i]=ta[i]=0;
      for (j=0;j<i;j++)
         if (a[j]>a[i]) t[i]++;
      for (j=i+1;j<n;j++)
         if (a[j]>a[i]) ta[i]++;
   }
   s=0;kk=(k*(k+1)/2)%1000000007;
   kk1=(k*(k-1)/2)%1000000007;
   for (i=0;i<n;i++)
      s=(s+t[i]*kk+ta[i]*kk1)%1000000007;
   printf("%ld",s);
   return 0;
}