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 boost() ios_base :: sync_with_stdio(0); cin.tie(); cout.tie(); cout << fixed; cout << setprecision(15); srand(time(NULL))
#define all(x) x.begin(), x.end()
using namespace std;

const int mod = (int)1e9 + 7;
int n, k, res = 0;
int a[2010];
int x[2010];
int y[2010];

int add (int i, int j) {
  return (i + j) % mod;
}

int mul (int i, int j) {
  return (i * j) % mod;
}

signed main () {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (a[i] > a[j]) {
        x[i]++;
        y[i] += (i < j);
      }
    }
    // cout << x[i] << " " << y[i] << endl;
  }
  for (int i = 0; i < n; i++) {
    // res = res + x[i] * k * (k - 1) / 2 + y[i] * k;
    int sum = (k  * (k - 1) / 2) % mod;
    res = add (res, mul (x[i], sum));
    res = add (res, mul (y[i], k));
    // cout << res << endl;
  }
  cout << res << endl;
  return 0;
}