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 (1.15.1) (judge verdict: Accepted)

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

fn read_vec2<T: std::str::FromStr>(n: usize) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}

fn read_col<T: std::str::FromStr>(n: usize) -> Vec<T> {
    (0..n).map(|_| read()).collect()
}

fn main() {
    let nk: Vec<usize> = read_vec();
    let n = nk[0];
    let k = nk[1] as u64;
    let mut a: Vec<u64> = read_vec();
    let mut b: Vec<u64> = vec![0; n];

    for i in 0..n {
        let mut cnt: u64 = 0;
        let tmp: u64 = a[i];
        for j in (i + 1)..n {
            if tmp > a[j] {
                cnt += 1;
            }
        }
        b[i] = cnt;
    }

    let mut b_sum: u64 = 0;

    for tmp in b {
        b_sum = (b_sum + tmp) % 1000000007;
    }

    a.sort();

    let mut c_sum: u64 = 0;

    let mut tmp = a[0];
    let mut cnt: u64 = 0;
    for i in 1..n {
        if tmp != a[i] {
            tmp = a[i];
            cnt = i as u64;
        }
        c_sum = (c_sum + cnt) % 1000000007;
    }

    let tmp1 = (k * b_sum) % 1000000007;
    let tmp2 = ((((k * (k - 1)) / 2) % 1000000007) * c_sum) % 1000000007;
    let ans = (tmp1 + tmp2) % 1000000007;
    println!("{}", ans);
}