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 Rust (1.42.0) (judge verdict: Wrong Answer)

fn main() {
    proconio::input! {
        n: usize,
        k: usize,
        a_s: [usize; n],
    }
    let mut tentosu = vec![(0usize, 0usize); n];
    for i in 0..n {
        for j in 0..i {
            if a_s[i] > a_s[j] {
                tentosu[i].0 += 1;
            }
        }
        for j in i + 1..n {
            if a_s[i] > a_s[j] {
                tentosu[i].1 += 1;
            }
        }
    }
    const M: usize = 1_000_000_007;
    let mut count = 0usize;
    for i in 0..n {
        count += tentosu[i].0 * (k - 1) * k / 2;
        count += tentosu[i].1 * (k + 1) * k / 2;
        count %= M;
    }
    println!("{}", count);
}