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)

#define rep(i, n) for(int i = 0; i < (int)n; i++)
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

int main()
{
#ifdef LOCAL
    // 以降 cin の入力元が 'xxx.txt' になる
    std::ifstream in("b.txt");
    std::cin.rdbuf(in.rdbuf());
#endif

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

    ll a[2000];

    rep (i ,n)
    {
        cin >> a[i];
    }

    ll ans = 0;
    rep (i, n)
    {
        ll cnt_self = 0;
        ll cnt_o = 0;
        for (ll self_i = i+1; self_i < n; self_i++)
        {
            if (a[i] > a[self_i])
            {

                cnt_self++;
                //cout << "self " << "a[" << i << "]:" << a[i] << " > " << a[self_i] << " cnt:" << cnt_self <<endl;
            }
        }
        for (ll self_i = 0; self_i < n; self_i++)
        {
            if (a[i] > a[self_i])
            {
                cnt_o++;
                //cout << "other " << "a[" << i << "]:" << a[i] << " > " << a[self_i] << " cnt:" << cnt_o <<endl;
                //cout << a[i] << " " << a[self_i] << endl;

            }
        }
        ans += (k*cnt_self) + (cnt_o*(k*(k-1)/2));

    }

    cout << ans%(1000000000LL+7);

    return 0;
}