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 int long long
#define rep(i, n) for (int i = (int)(0); i < (int)(n); ++i)
#define reps(i, n) for (int i = (int)(1); i <= (int)(n); ++i)
#define rrep(i, n) for (int i = ((int)(n)-1); i >= 0; i--)
#define rreps(i, n) for (int i = ((int)(n)); i > 0; i--)
#define irep(i, m, n) for (int i = (int)(m); i < (int)(n); ++i)
#define ireps(i, m, n) for (int i = (int)(m); i <= (int)(n); ++i)
#define SORT(v, n) sort(v, v + n);
#define REVERSE(v, n) reverse(v, v+n);
#define vsort(v) sort(v.begin(), v.end());
#define all(v) v.begin(), v.end()
#define mp(n, m) make_pair(n, m);
#define cout(d) cout<<d<<endl;
#define coutd(d) cout<<std::setprecision(10)<<d<<endl;
#define cinline(n) getline(cin,n);
#define replace_all(s, b, a) replace(s.begin(),s.end(), b, a);
#define PI (acos(-1))
#define FILL(v, n, x) fill(v, v + n, x);
#define sz(x) int(x.size())

using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

const int INF = 1e9;
const int MOD = 1e9+7;
const ll LINF = 1e18;

long long modpow(long long a, long long n, long long mod) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = res * a % mod;
    (a *= a) %= mod;
    n >>= 1;
  }
  return res;
}

const int maxf = 10000000;
long long fact[maxf+5];

void fact_init(int n, long long mod){
  fact[0] = 1;
  for(int i=1; i<=n; i++){
    fact[i] = (fact[i-1] * i) % mod;
  }
}

long long comb(long long a, long long b, long long mod){
  if(a<b || a<0 || b<0) return 0;
  if(fact[0] == 0) fact_init(maxf, mod);
  return (fact[a] * modpow((fact[a-b]*fact[b])%mod, mod-2, mod)) % mod;
}

signed main()
{
  cin.tie( 0 ); ios::sync_with_stdio( false );
  ll n,k; cin>>n>>k;
  vll a(n);
  rep(i,n) cin>>a[i];
  ll cnt=0;
  rep(i,n){
    irep(j,i+1,n){
      if(a[i]>a[j]) cnt++;
    }
  }
  ll cnt2=0;
  rrep(i,n){
    for(int j=i-1; j>=0; j--){
      if(a[j]<a[i]) cnt2++;
    }
  }
  ll ans=0;
  ans=((cnt*((k+1)*k/2%MOD)) + cnt2*((k*(k-1)/2%MOD)))%MOD;
  cout<<ans<<endl;
}