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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <algorithm>
#include <cstdio>

using namespace std ;

typedef long long LL ;
const int N=2e3+10 ;
const LL P=1e9+7 ;

int a[N] ;
int n ;
LL k ;

int main ()
{
	int i , j ;
	scanf("%d%lld",&n,&k);
	for ( i=1 ; i<=n ; i++ ) scanf("%d",a+i);
	LL res=0 ;
	for ( i=1 ; i<=n ; i++ )
	{
		LL s=0 , t=0 ;
		for ( j=1 ; j<i ; j++ )
			if ( a[j]>a[i] ) s++;
		for ( j=1 ; j<=n ; j++ )
			if ( a[j]>a[i] ) t++;
		res=(res+k*(k-1)/2ll%P*t%P+k*s%P)%P;
	}
	printf("%lld\n",res);
}