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<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f3f;
const int mod = 1e9+7;
const int maxn = 1e5+500;
ll T,n,m;
int a[maxn];
ll num[maxn];
ll pos[maxn];
int main() {
    cin>>n>>m;
    int cnt = 0;
	for(int i=0;i<n;i++){
		scanf("%d",&a[i]);
	}
	for(int i=0;i<n;i++){
		for(int j=i+1;j<n;j++){
			if(a[i]>a[j]){
				num[i]++;
			}
		}
	}
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			if(a[i]>a[j]){
				pos[i]++;
			}
		}
	}
	ll sum=0;
	for(int i=0;i<n;i++){
		sum+=pos[i];
	}
	
	sum%=mod;
	ll ssum = 0;
	ll temp = (m-1)*m/2;
	temp%=mod;
	temp*=sum;
	temp%=mod;
	ssum = temp;
	ll p = 0;
	for(int i=0;i<n;i++){
		p+=num[i];
	}
	p%=mod;
	p*=m;
	p%=mod;
	ssum+=p;
	ssum%=mod;
	cout<<ssum<<endl;
    return 0;
}