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: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i,a,b) for(int i=int(a);i<int(b);++i)
#define SIZE 200005
#define INF 1000000005LL
#define MOD 1000000007
using namespace std;
typedef long long int ll;
typedef pair <int,int> P;
int main(){
    ll n;
    ll k;
    cin >> n>>k;
    vector<ll> loop(n);
    map<ll,pair<ll,ll>> data;
    rep(i,0,n){
        cin >> loop[i];
    }
    rep(i,0,n){
        rep(j,0,n){
            if(loop[i]<loop[j]&&i>j){
                data[i].first++;
                data[i].second++;

            }else if(loop[i]<loop[j]&&i<j){
                data[i].second++;
            }
        }
    }
    //cout<<data[1].first<<data[1].second<<endl;
    ll ans=0;
    rep(i,0,n){
        ans+=(data[i].first*k)%MOD;
        ans%=MOD;
        ans+=(data[i].second*(k-1)%MOD*k/2)%MOD;
        ans%=MOD;
    }
    cout<<ans<<endl;


  return 0;
}