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)

#include<bits/stdc++.h>
#include<stdio.h>
using namespace std;
typedef int64_t ll;
typedef long double ld;

void solve() {
    const ll M = 1000000007;
    ll n, k;
    cin >> n >> k;
    ll ans = 0, peeche[n] = {0}, aage[n] = {0}, a[n];
    for (ll i = 0; i < n; i++) cin >> a[i];
    for (ll i = 0; i < n; i++) {
        for (ll j = 0; j < n; j++) {
            if (j != i) {
                if (j < i && a[i] > a[j]) peeche[i]++;
                else if (j > i && a[i] > a[j]) aage[i]++;
            }
        }
    }

    ll kf = ((k * (k + 1)) / 2) % M, kb = ((k * (k - 1)) / 2) % M;
    for (ll i = 0; i < n; i++) {
        ans += (((peeche[i] * kb) % M) + ((aage[i] * kf) % M)) % M; 
    }
    cout << ans;
}

int main() {
    ios_base :: sync_with_stdio(false); 
    cin.tie(NULL);
    ll t = 1;
    //cin >> t;
    while(t--) {
        solve();
    }
    return 0;
}