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>

#define SIZE 2000
#define MOD  1000000007

#define ll        long long
#define rep(i, n) for(ll i=0; i<n; ++i)


int main() {
    ll  n, k;
    ll  a[SIZE];
    ll  ans = 0LL;
    ll  count[SIZE] = { 0 };
    std::cin >> n >> k;
    rep(i, n) std::cin >> a[i];

    rep(i, n) {
        rep(j, n) {
            if (a[i] > a[j]) ++count[i];
        }
    }

    rep(i, n-1) {
        ll tmp = 0LL;
        for (ll j=i+1; j<n; ++j) {
            if (a[i] > a[j]) ++tmp;
        }
        ans += tmp * k;
    }

    rep(i, n) {
        ll tmp = count[i] * ( (k-1) * k / 2);
        tmp %= MOD;
        ans = (ans + tmp) % MOD;
    }
    std::cout << ans << std::endl;

    return 0;
}