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 int long long
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define F first
#define S second

using namespace std;
typedef pair<int,int> P;

const int MOD=1000000007;
int INF=100100100100100;

int a[2010];
int zen[2010];
int po[2010];

int add(int a,int b){
  return (a+b)%MOD;
}
int prod(int a,int b){
  return (a*b)%MOD;
}
    
signed main(){
  int n,k;
  cin>>n>>k;
  rep(i,n)cin>>a[i];
  rep(i,n){
    rep(j,n){
      if(a[j]<a[i])zen[i]++;
    }
  }
  rep(i,n){
    for(int j=i+1;j<n;j++){
      if(a[j]<a[i])po[i]++;
    }
  }
  int ans=0;
  rep(i,n){
    int m=prod(zen[i],k*(k-1)/2);
    ans=add(ans,add(prod(po[i],k),zen[i]*k*(k-1)/2));
  }
  cout<<ans<<endl;
  return 0;
}