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 <iomanip>
#include <cmath>
#include <string>
#include <vector>
#include <cstdlib>
#include <map>
#include <queue>
#include <algorithm>
#include <stack>
#include <functional>
#define rep(i,j,n) for(i=j;i<(n);i++)
#define rrep(i,j,n) for(i=j;i>(n);i--)
typedef long long int lli;
#define N 1000000007

using namespace std;

//lli gcd(lli a,lli b){
//    lli x,y;
//    x=max(a,b);
//    y=min(a,b);
//    if(x%y==0){
//        return y;
//    }
//    else{
//        return gcd(y,x%y);
//    }
//}
//
//lli lcm(lli a,lli b){
//    lli x,y;
//    x=max(a,b);
//    y=min(a,b);
//    return x*y/gcd(x,y);
//}

int main() {
	lli n, k, a, ans = 0, b = 0;
	lli i, j;
	vector<lli> vec;
	cin >> n >> k;
	rep(i, 0, n) {
		cin >> a;
		vec.push_back(a);
	}
	rep(i, 0, n - 1) {
		rep(j, i + 1, n) {
			if (vec[i] > vec[j]) {
				ans++;
				ans %= N;
			}
		}
	}
	rep(i, 0, n) {
		rep(j, 0, n) {
			if (vec[i] > vec[j]) {
				b++;
			}
		}
	}
	//    cout<<ans<<" "<<b<<endl;
	ans *= k;
	ans %= N;
	k = ((k - 1) * k) / 2 % N;
	//    cout<<b*(k/2)<<endl;
	ans += b * k % N;
	ans %= N;
	cout << ans << endl;
}