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)

// Author -> Preeyadarshee Dev
//
// train hard win easy :D
// nik is love
// nik is motivation
//
// do modulo operations very carefully
// try to avoid binary search, use linear search preferrably to make calculations easier
// always make your custom binary search and watch out for overflows
// read the problems carefully
// never rely on floating point calculation
#include<bits/stdc++.h>
#define mp                    make_pair
#define pb                    push_back
#define ll                    long long
#define ss                       second
#define ff                        first
#define fr(i,a,b)   for(ll i=a;i<b;i++)
#define f(i,n)                fr(i,0,n)
#define rf(i,b,a) for(ll i=b-1;i>=a;i--)
#define r(i,n)                rf(i,n,0)
#define inf                   1000000007
#define eps                   0.000000000000001
#define sz(a)                int((a).size())
#define all(c)               (c).begin(),(c).end()
#define tr(c,i)              for(typeof(c).begin() i = (c).begin(); i != (c).end(); i++)
#define present(c,x)         ((c).find(x) != (c).end())
#define cpresent(c,x)        (find(all(c),x) != (c).end())
#define pi                   pair<int,int>
#define pll                  pair < ll , ll >
#define endl                 '\n'
#define PI                   3.141592653589793238462643383279502884197169399375105820974944592307816406286
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
typedef vector< int > vi;
typedef vector< vi > vvi;
int dx[8]={1,-1,0,0,-1,-1,1,1};
int dy[8]={0,0,-1,1,-1,1,-1,1};
vector<int> adj[200010],vis(200010,0),dis(200010,inf),fin(200010,0),col(200010,0),par(200010,0),v;
set<int> st[100010];
int myrandom (int i) { return std::rand()%i;}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    //ifstream fin;
    //ofstream fout;
    //fin.open("input.txt");
    //fout.open("output.txt");
    ll n,k;
    cin>>n>>k;
    vector<ll> v(n);
    f(i,n)
    {
        cin>>v[i];
    }
    ll cnt1=0,i=0,cnt=0;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<=i;j++)
        {
            if(v[j]>v[i])
            {
                cnt++;
            }
        }
    }
    sort(all(v),greater<ll>());
    ll temp=0;
    while(i<n)
    {
        ll num=v[i];
        while(i<n&&num==v[i])
        {
            i++;
            temp++;
        }
        cnt1=cnt1+(n-temp);
    }
    ll num=(k*(k-1))/2;
    num=num%inf;
    num=(num*(cnt1%inf))%inf;
    ll ans=(((cnt%inf)*(k%inf))%inf+num)%inf;
    cout<<ans<<endl;
}