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>

#define NREP() for(ll i = 0; i < n; i++)
#define REP(i, n) for(ll i = 0; i < n; i++)
#define ALL(x) (x).begin(), (x).end()
#define MSG(x) cout << x << endl;
#define YN(x) x ? cout << "YES" << endl : cout << "NO" << endl;
#define Yn(x) x ? cout << "Yes" << endl : cout << "No" << endl;
#define yn(x) x ? cout << "yes" << endl : cout << "no" << endl;

using namespace std;
using ll = long long;
const static ll MOD = 1e9 + 7;

ll multi_mod(ll a, ll b) {
    return (a%MOD * b%MOD)%MOD;
}

int main(int argc, char *argv[]){
    ll n, k;
    cin >> n >> k;

    ll a[n+10];
    ll s[n+10];
    NREP() {
        cin >> a[i];
        s[i] = a[i];
    }
    sort(s, s+n);
    ll count_reverse = 0;

    for (int i = 0; i < n-1; ++i) {
        for (int j = i; j < n; ++j) {
            if (a[i] > a[j]) count_reverse++;
        }
    }

    ll min_sum = 0;
    ll p = s[0];
    ll prev = 0;

    for (int i = 0; i < n; ++i) {
        if (p < s[i]) {
            min_sum += i;
            prev = i;
            p = s[i];
        } else if (p == s[i]) {
            min_sum += prev;
        }
    }

    cout << count_reverse << " " << min_sum << endl;
    count_reverse %= MOD;
    min_sum %= MOD;


    if (k == 2) {
        cout << (multi_mod(count_reverse, k) + min_sum) % MOD << endl;
    } else {
        cout << (multi_mod(count_reverse, k) + multi_mod(min_sum, (multi_mod(k-1, k) / 2))) % MOD << endl;
    }
}