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)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
using P   = pair<ll, ll>;
using mat = vector<vector<ll>>;
typedef string::const_iterator State;
class ParseError {};

void IOS() { ios::sync_with_stdio(false), cin.tie(0); }

constexpr ll INF = 1e16;
constexpr ll INM = 114514;
constexpr ll MOD = 1e9 + 7;
constexpr ld EPS = 1e-12;

template <typename T>
void dump(T x) { cout << x << endl; }
void dumpf(ld x, int t) { cout << setprecision(t) << fixed << x << endl; }
template <typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val) {
	fill((T *)array, (T *)(array + N), val);
}
inline ll mod(ll a, ll b) {
	if (a > 0) return a % b;
	if (a % b == 0) return 0;
	ll x = -a / b + 1;
	a += x * b;
	return a % b;
}
ll powm(ll a, ll b, ll c) {
	if (b == 0)
		return 1;
	else if (b % 2 == 0) {
		ll d = powm(a, b / 2, c);
		return (d * d) % c;
	} else
		return (a * powm(a, b - 1, c)) % c;
}
ll gcd(ll a, ll b) {
	if (b == 0) return a;
	return gcd(b, a % b);
}
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
ll dx[4] = {1, 0, -1, 0};
ll dy[4] = {0, 1, 0, -1};

int main() {
	IOS();
	ll n, k;
	cin >> n >> k;
	vector<ll> a(n), pre(n, 0), pro(n, 0);
	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[j] < a[i]) pre[i]++;
		}
		for (int j = i + 1; j < n; j++) {
			if (a[j] < a[i]) pro[i]++;
		}
	}
	ll ans = 0;
	for (int i = 0; i < n; i++) {
		ll k1 = ((k + 1) * k / 2) % MOD;
		ll k2 = ((k - 1) * k / 2) % MOD;
		ans += pre[i] * k2;
		ans %= MOD;
		ans += pro[i] * k1;
		ans %= MOD;
		//printf("%d %d\n", pre[i], pro[i]);
	}
	dump(ans);
	return 0;
}