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 <algorithm>
#include <vector>

using namespace std;

int main()
{
    unsigned long long N, K;
    cin >> N >> K;
    vector<int> A(N);
    for(int i=0;i<N;i++){
        cin >> A[i];
    }

    unsigned long long count, count_same, sum_count = 0;
    for(int i=0;i<N;i++){
        count = 0;
        for(int j=i+1;j<N;j++){
            if(A[i] > A[j]) count++;
        }
        sum_count += count;
    }

    sum_count *= K;
    sum_count += ((K * (K-1)) / 2) * ((N * (N-1)) / 2);

    sort(A.begin(), A.end());

    unsigned long long pre = -1;
    for(int i=0;i<N;i++){
        if(pre == A[i]) continue;
        count_same = 0;
        for(int j=i+1;j<N;j++){
            if(A[i] == A[j]){
                count_same++;
                pre = A[i];
            }
        }
        count_same = (count_same * (count_same+1)) / 2;
        sum_count -= (count_same * ((K * (K-1)) / 2));
    }

    cout << sum_count % (1000000007) << endl;
}