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 <iostream>
#include <math.h>
#include <string>
#include <string.h>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <stdlib.h>
#include <deque>  
#include <iomanip>

using namespace std;

#define ll long long
#define INF 1e9
const ll MOD = 1e9 + 7;
#define rep(i, n) for(int i = 0; i < n; i++)
#define loop(i, a, n) for(int i = a; i < n; i++)
#define all(in) in.begin(), in.end()
int main(){
    int N, K;
    cin >> N >> K;
    vector<int> a(N);
    rep(i, N) cin >> a[i];
    ll ans = 0;
    for (int i = 0; i < N; i++) {
        ll count = 0;
        ll count_low = 0, count_upper = 0;
        for (int j = 0; j < N; j++) {
            if (a[i] > a[j]){
                if (j < i) count_low++;
                else count_upper++;
            }
        }
        ll sum = 0;
        if (count_low == 0){
            sum = (count_upper + count_upper * K ) * K / 2 % MOD;
        }else{
            sum = (count_upper + count_upper + (count_upper + count_low) * (K - 1)) * K /2  % MOD;
        }
        ans += sum;
        ans %= MOD;
    }
    cout << ans % MOD << endl;
    return 0;
}