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: Wrong Answer)

#include <iostream>
#include <stack>
#include <algorithm>
#include <vector>
#include <math.h>
#include <stdio.h>
#include <functional>
#include <string>
#include <cstdlib>
#include <numeric>
#include <cstdbool>
#include <map> 
#include<set>
#include<queue>

typedef long long ll;


#define rep(i,n) for(ll i=0;i<n;i++)
using namespace std;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

const long long INF = 1000000000000000;
const ll inf = -1e18;
typedef pair<ll, ll> P;
ll ma = 1000000000 + 7;
ll mx = 1000003;
ll h, w, n,m,k; string s,t;
char maze[60][60];
int dis[101][101]; int graph[52][52]; bool vis[52];
int dx[4] = { 1, 0, -1, 0 };
int dy[4] = { 0, 1, 0, -1 };
ll gcd(ll x, ll y) {
	if (x % y == 0) return y;
	return gcd(y, x % y);

}
ll lcm(ll a,ll b) {
	ll g = gcd(a, b);
	return a / g * b;
}
void comb(vector<vector <ll> >& v) {
	for (ll i = 0; i < v.size(); i++) {
		v[i][0] = 1;
		v[i][i] = 1;
	}
	for (ll k = 1; k < v.size(); k++) {
		for (int j = 1; j < k; j++) {
			v[k][j] = (v[k - 1][j - 1] + v[k - 1][j])%ma;
		}
	}
}

ll GetDigit(ll num) {
	return log10(num) + 1;
}

int main() {
	cin >> n >> k;
	vector<ll> a(n);
	ll cnt = 0;
	rep(i, n) {
		cin >> a[i];
	}
	rep(i, n) {
		ll pp = a[i];
		ll rcnt = 0;
		ll lcnt = 0;
		rep(j, n) {
			if (i != j) {
				if (i>j&&pp > a[j]) {
					lcnt++;
				}
				else if (i<j && pp > a[j]) {
					rcnt++;
				}
			}
		}
		ll kcnt = (lcnt + rcnt) * k*(k-1) / 2 % ma;
		cnt += kcnt;
		cnt %= ma;
		cnt += k*rcnt;
		cnt %= ma;
	}
	cout << cnt << endl;
}