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: Accepted)

#include <bits/stdc++.h>

using namespace std;

#define rep(i, n) for(int i = 0; i < (n); ++i)
#define eee(a) (cerr << (#a) << ": " << (a) << endl);
#define ppp(a) (cout << (a) << endl);
#define cop(a) for(auto x:(a)){cout << x << endl;}


using ll = long long;
using pint = pair<int, int>;
using pll = pair<ll, ll>;
using mint = map<int, int>;

int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};

const char sp = ' ';
const char cmm = ',';
const int MOD = 1e9+7;
const int INF = 1e9;
const ll LINF = 1e18;

ll mod(ll a, ll b){return (a%b+b)%b;}
ll gcd(ll a, ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b){return a*b/gcd(a,b);}
void Yes(){cout << "Yes" << endl;}
void No(){cout << "No" << endl;}
void Judge(bool b){b?Yes():No();}
void YES(){cout << "YES" << endl;}
void NO(){cout << "NO" << endl;}
void JUDGE(bool b){b?YES():NO();}
ll powmod(ll b, ll e){ll r=1;while(e>0){if(e&1)r=(r%MOD)*(b%MOD)%MOD;b=(b%MOD)*(b%MOD)%MOD;e>>=1;}return r;}
double distance(ll x1, ll y1, ll x2, ll y2){return sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));}

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template<typename T1, typename T2> ostream& operator<<(ostream& s, const pair<T1, T2>& p) {return s << "(" << p.first << ", " << p.second << ")";}
template<typename T> ostream& operator<<(ostream& s, const vector<T>& v) {
    int len = v.size();
    s << '[';
    for (int i = 0; i < len; ++i) {
        s << v[i]; if (i < len-1) s << ", ";
    }
    s << ']';
    return s;
}

struct aaa{
    aaa(){
        // comment this out when you use scanf or printf
        // cin.tie(0); ios::sync_with_stdio(0);
        cout<<fixed<<setprecision(15);
    };
}fastio;

int main(){

    ll n, k;
    cin >> n >> k;

    vector<ll> a(n);
    rep(i, n) scanf("%lld", &a[i]);
    vector<ll> b(a);
    for(ll x : a){
        b.emplace_back(x);
    }

    ll inv = 0;
    ll inv2 = 0;
    for(int i = 0; i < 2*n; ++i){
        for(int j = i+1; j < 2*n; ++j){
            if(b[i] > b[j] and (i < n and j >= n)) ++inv2;
            else if(b[i] > b[j] and j < n) ++inv;
        }
    }
    inv %= MOD;
    inv2 %= MOD;
    // eee(inv);
    // eee(inv2);
    ll x;
    if(k%2){
        x = (k-1)/2*k%MOD;
    }
    else{
        x = k/2*(k-1)%MOD;
    }
    ll ans = (k*inv%MOD + x*inv2%MOD)%MOD;
    ppp(ans);

    return 0;
}