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 <iostream>
#include <string>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <vector>
#include <cmath>
#include <math.h>
#include <algorithm>
#include <map>
#include <sstream>
#include <queue>
#include <deque>
#include <set>
#include <stack>
#include <functional>
#include <utility>
#include <stdio.h>
#define ll long long
#define MP make_pair
#define PII pair<int, int>
#define PLL pair<ll, ll>
#define PLI pair<ll, int>
#define PIL pair<int, ll>
#define F first
#define S second
#define SHN 1e9
#define LSHN 1e18
using namespace std;
ll n,m;
ll a[2010]; 
ll b[2010][2];
ll res=0;
ll md=1e9+7;
int main()
{
	cin >>n>>m;
	for (int i=0;i<n;i++) cin >>a[i];
	for (int i=0;i<n;i++)
	{
		for (int j=0;j<i;j++) if (a[i]>a[j]) b[i][0]++;
		for (int j=i+1;j<n;j++) if (a[i]>a[j]) b[i][1]++;
	}
	for (int i=0;i<n;i++)
	{
		//cout <<i<<" "<<b[i][0]<<" "<<b[i][1]<<" "<<res<<" ##\n";
		ll r=(m+1)*m/2;
		r%=md;
		r=(r*b[i][1])%md;
		res=(res+r)%md;
		r=m*(m-1)/2;
		r%=md;
		r=(r*b[i][0])%md;
		res=(res+r)%md;
	}
	cout <<res;
	return 0;
}