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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

long int_compare(const long* a, const long* b)
{
    return (*a > *b) ? 1
                     : (*a < *b) ? -1
                                 : 0;
}

void copy_arr(long* dst, long* src, long n)
{
    for (long i = 0; i < n; i++) {
        *dst++ = *src++;
    }
}

long get_index(long* a, long key, long n)
{
    long past = a[0];
    long order = 0;
    for (long i = 0; i < n; i++) {
        if (past != a[i]) {
            order++;
        }
        if (a[i] == key) {
            return order;
        }
        past = a[i];
    }
    return 0;
}

int main(void)
{
    const long mod = 1000000007;

    long n, k;
    scanf("%ld %ld", &n, &k);

    long* a = calloc(n, sizeof(long));
    for (long i = 0; i < n; i++) {
        scanf("%ld", &a[i]);
    }

    long count = 0;

    long* sorted = calloc(n, sizeof(long));
    copy_arr(sorted, a, n);
    qsort(sorted, n, sizeof(long), (int (*)(const void*, const void*))(*int_compare));

    for (long i = 0; i < n; i++) {
        long first_elem = 0;
        for (long j = i + 1; j < n; j++) {
            if (a[i] > a[j]) {
                first_elem++;
            }
        }
        long idx = get_index(sorted, sorted[i], n);
        count += (k * (2 * first_elem + (k - 1) * idx) / 2) % mod;
        count %= mod;
    }

    printf("%ld\n", count);
}