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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#include <math.h>
#include <iomanip>
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const int INF=1001001001;
vector<pair<int64_t,int64_t>>prime_factorize(int64_t x){
  vector<pair<int64_t,int64_t>>p;
  for(int64_t i=2;i*i<=x;i++){
    int cnt=0;
    if(x%i==0){
    while(x%i==0){cnt++;x/=i;}
    p.push_back(make_pair(i,cnt));
    }
  }
  if(x!=1){p.push_back(make_pair(x,1));}
  return p; 
}

int main() {
  int64_t N,K;
  cin>>N>>K;
  vector<int>A(N);
  for(int i=0;i<N;i++){
    cin>>A[i];
  }
  int64_t p=1000000007;
  int64_t sum=0;
  int64_t cnt=0;
  for(int i=0;i<N;i++){
    for(int j=i+1;j<N;j++){
      if(A[i]>A[j]){cnt++;}
    }
  }
  sum+=(cnt*K%p);
  cnt=0;
  for(int i=0;i<N;i++){
    for(int j=0;j<N;j++){
      if(A[i]>A[j]){cnt++;}
    }
  }
  sum+=cnt%p*(K*(K-1)/2%p);//割り算のMODは途中でMODしてはいけない
  cout<<sum%p<<endl;
  return 0;
}