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: Wrong Answer)

#[allow(dead_code)]
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()
}
#[allow(dead_code)]
fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}
#[allow(dead_code)]
fn read_vec2<T: std::str::FromStr>(n: u32) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}
#[allow(dead_code)]
fn yn(result: bool) {
    if result {
        println!("Yes");
    } else {
        println!("No");
    }
}

fn main() {
    let v = read_vec::<usize>();
    let n = v[0];
    let k = v[1] as u64;

    let a = read_vec::<u64>();

    let mut at_count = 0;
    let mut top_count = 0;
    for i in 0..n-1 {
        for j in i+1..n {
            if a[i] > a[j] {
                at_count += 1;
            } else if i == 0 && a[i] < a[j] {
                top_count += 1;
            }
        }
    }

    let ks:u64 = k * (k + 1) / 2;// % 1000000007;
    //println!("{}", ks * at_count % 1000000007);
    let ts:u64 = (k-1) * k / 2;
    //println!("{}", ts * top_count % 1000000007);

    //println!("{} {} {} {}", ks, at_count, ts, top_count);
    let mut ans: u64 = 0;
    for _ in 0..at_count {
        ans += ks;
        ans %= 1000000007;
    //    println!("{}", ans);
    }
    for _ in 0..top_count {
        ans += ts;
        ans %= 1000000007;
//        println!("{}", ans);
    }
    //println!("{} {}", ks, at_count);
    println!("{}", ans);



    /* loop pattern
    let mut count = 0;
    loop {
        if  { break; }
        count += 1;
    }
    println!("{}", count);
    */

    /* yes/no pattern
    yn(a == b);
    */
}