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 <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<ll,ll> pll;
typedef vector<bool> vb;
const ll oo = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-9;
#define sz(c) ll((c).size())
#define all(c) begin(c), end(c)
#define FOR(i,a,b) for (ll i = (a); i < (b); i++)
#define FORD(i,a,b) for (ll i = (b)-1; i >= (a); i--)
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define xx first
#define yy second
#define TR(X) ({ if(1) cerr << "TR: " << (#X) << " = " << (X) << endl; })

const ll MOD = 1e9 + 7;

// 2 1 | 2 1 | 2 1 -> 3 + 2 + 1
// 1 2 | 1 2 | 1 2 -> 2 + 1


int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	ll n, k;
	cin >> n >> k;

	vl a(n);
	FOR(i,0,n) cin >> a[i];

	ll res = 0;
	FOR(i,0,n) FOR(j,0,n) if (a[i] > a[j]) {
		ll cnt = k*(k+1)/2;
		if (i > j) cnt -= k;
		res = (res + cnt) % MOD;
	}
	cout << res << endl;

}