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>
using namespace std;
typedef long long ll;
const ll mod=1e9+7;
const ll INF=mod*mod;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define all(x) (x).begin(),(x).end()
ll min(ll a,ll b){
  if(a>b){
    return b;
    }
      else{
        return a;
      }
  }
  
  ll max(ll a,ll b){
    if(a>b){
      return a;
    }
    else{
      return b;
    }
  }

ll power(ll a,ll b){
 ll ans=1;
 for(ll i=1;i<=b;i++){
ans*=a;
  }
  return ans;
}

 ll gcd(ll m,ll n){
 	if(n==0){
       return m;
     }
   return gcd(n,m%n);
 }
ll lcm(ll a,ll b){
  ll g=gcd(a,b);
  return a/gcd(a,b)*b;
		

}
int main(){
  
  ll N,K;
  cin>>N>>K;
  vector<ll>A(N);
  vector<ll>C(N);
  
  rep(i,N){
    cin>>A[i];
    C[i]=A[i];
    }
  ll ans=0;
  
  
  rep(i,N){//現在の回転範囲
    vector<ll>a(N-i);//a[i]の後のうちこれより小さいものの数
  
    for(ll j=i;j<N;j++){
     a.push_back(A[j]);
    }
    sort(all(a));
      
    ans+=(lower_bound(a.begin(),a.end(),A[i])-a.begin())*K;
  ans%=mod;
  }
  ll ans2=0;
 sort(all(C));//現在の後の場所
  rep(i,N){
 
 ans2+=(lower_bound(all(C),C[i])-C.begin())*K*(K-1)/2;
    ans2%=mod;
  }

  


  cout<<(ans+ans2)%mod<<endl;
}