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 int ll;
typedef pair<int,int> P;
typedef pair<ll,ll> Pll;
typedef pair<string,string> Pstring;
typedef pair<double,double> Pdouble;


#define REP(i,n) for(ll i=0;i<n;i++)
#define REP1(i,n) for(ll i=1;i<=n;i++)
#define Precision13 cout << fixed << setprecision(13)
const double PI=3.14159265358979323846;
const int MAX = 510000;
const int MOD = 1000000007;
const int INF = 1<<29;
using Graph = vector<vector<int>>;

int main() {
  ll n,k;
  cin >> n >> k;
  vector<ll> a(n);
  REP(i,n) cin >> a.at(i);

  ll s1=0;
  ll s2=0;
  for(int i=0; i<n-1;i++){
    for(int j=i+1;j<n;j++){
      if(a[i]>a[j]){
        s1++;
      }else if(a[i]<a[j]){
        s2++;
      }
    }
  }

  ll ans;
  ans = k*(k+1)/2%MOD*s1%MOD+(k-1)*(k)/2%MOD*s2%MOD;
  cout << ans%MOD << endl;

}