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>
#define int long long
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define DEC(i, a, b) for (int i = (a); i > (b); --i)
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define pb push_back
#define ALL(obj) (obj).begin(), (obj).end()
#define debug(x) cerr << #x << ": " << x << '\n'
using namespace std;
typedef long long ll;
const int INF = (int)1e9;
const int MOD = (int)1e9 + 7;
const ll LINF = (int)1e18;
const double EPS = 1e-9;

signed main(){
  cin.tie(nullptr);
  ios::sync_with_stdio(false);

  int n,k;
  cin >> n >> k;

  int ans = 0;
  int ans_ = 0;
  vector<int> v;
  REP(i,n) {
    int a_;
    cin >> a_;
    for(auto e: v) {
      if (e > a_) ++ans_;
    }
    v.pb(a_);
  }
  ans += (ans_ * k) % MOD;

  if (k > 1) {
    int rec = 0;
    for (auto e: v) {
      for (auto f: v) {
        if (f > e) ++rec;
      }
    }

    if (k > 2) ans += rec * ((k-1)*(k)/2 %MOD) %MOD;
    else ans += rec %MOD;
  }
  ans %= MOD;
  cout << ans << endl;

  return 0;
}