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 <string>
#include <map>
#include <queue>
#include <algorithm>
#include <vector>
#include <math.h>
#include <bitset>
#include <cmath>
#include <set>

#define rep(i,s,n)for(int i = s;i<n;i++)
#define repe(i,s,n)for(int i = s;i<=n;i++)
#define rep_r(i,s,n)for(int i = n-1;i>=s;i--)
#define rep_re(i,s,n)for(int i = n;i>=s;i--)
using namespace std;
typedef long long ll;
static const ll MOD = 1e9 + 7;
static const ll INF = 1e25;
static const ll MAX_N = 100001;
static const ll MAX_M = 100001;

ll a[4001];
ll b[4001];
ll c[4001];
int main() {
	ll n, k; cin >> n >> k;
	rep(i, 0, n) {
		ll A; cin >> A;
		a[i] = A;
		a[i + n] = A;
	}

	rep(i, 0, n) {
		rep(j, i + 1, 2 * n) {
			if (j < n) {
				if (a[i] > a[j]) {
					b[i]++;
				}
			}
			else {
				if (a[i] > a[j]) {
					c[i]++;
				}
			}
		}
	}
	ll ans = 0;
	rep(i, 0, n) {
		ll B = (b[i] * k) % MOD;
		ll C = (c[i] * (k * (k - 1) / 2)) % MOD;
		ll A = (B + C) % MOD;
		ans = (ans + A) % MOD;
	}

	cout << ans << endl;
	return 0;
}