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 <bits/stdc++.h>
 
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n)  for(int i=0;i<(n);++i)
#define REPr(i, n) for(int i=(n)-1;i>=0; --i)
#define FORq(i, m, n) for(int i = (m);i <= (n);++i)
#define FORqr(i, m, n) for(int i = (n);i >=(m);--i)
#define PB push_back
#define MP make_pair
#define DEBUG printf("%s\n","debug")
#define fst first
#define snd second
#define SIN(x, S) (S.count(x) != 0)
#define M0(x) memset(x,0,sizeof(x))
#define FILL(x, y) memset(x,y,sizeof(x))
#define MM(x) memset(x,-1,sizeof(x))
#define ALL(x) (x).begin(),(x).end()
using namespace std;
typedef pair<int, int> PII;
typedef pair<long long, long long> PLL;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<long long> VL;
typedef long long ll;
typedef long long integer;
///////////////////////////////////////////////
/* (๑╹◡╹) */
ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll MOD = 1000000007;
template<class T>
inline bool chmin(T &a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
 /* (๑╹◡╹)(๑╹◡╹)(๑╹◡╹)*/
template<class T>
inline bool chmax(T &a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}
/* (๑╹◡╹)(๑╹◡╹)(๑╹◡╹)(๑╹◡╹)(๑╹◡╹)(๑╹◡╹) */
///////////////////////////////////////////////
const int MAX = 510000;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
 



int main() {
	ll ans = 0;
	ll N,K; cin >> N >> K;

	vector<ll> A;
	for (int i = 0; i < N; i++)
	{
		ll a; cin >> a;
		A.PB(a);
	}
	
	ll c = 0;//Aの中での転倒数
	for (int i = 0; i < N-1; i++)
	{
		for (int j = i+1; j < N; j++)
		{
			if(A[i]>A[j]){c++; c = c % MOD;
			}
		}
	}
	ll d = 0;//Aの外での転倒数
	for (int i = 0; i < N-1; i++)
	{
		for (int j = 0; j < N; j++)
		{
			if(A[i]>A[j] and i != j){d++; d = d % MOD;
			}
		}
	}
	c = (c * K) % MOD;
	d = ( ( ( K*(K-1)/2 ) % MOD ) * d ) % MOD;
	ans = ( c + d ) % MOD;
	ans = ans % MOD;
	cout << ans << endl;
	}