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 main() {
    let (n, k) = i::u2();
    let a = i::uv(n);
    const M: usize = 1_000_000_007;

    let mut all = vec![0; n];
    let mut after = vec![0; n];
    for i in 0..n {
        for j in 0..n {
            if a[i] > a[j] {
                all[i] += 1;
                if i < j {
                    after[i] += 1;
                }
            }
        }
    }
    let mut ans = 0usize;
    for i in 0..n {
        ans += after[i] * k % M + all[i] * k * (k - 1) / 2 % M;
    }
    println!("{}", ans % M);
    //dbg!(all, after);
}

#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::{HashMap, HashSet, VecDeque};
#[allow(unused_imports)]
use std::io::*;
#[allow(unused_imports)]
use std::num::*;
#[allow(unused_imports)]
use std::str::*;

#[allow(dead_code)]
mod i {
    use super::*;

    pub fn read<T: FromStr>() -> T {
        stdin()
            .bytes()
            .map(|c| c.unwrap() as char)
            .skip_while(|c| c.is_whitespace())
            .take_while(|c| !c.is_whitespace())
            .collect::<String>()
            .parse::<T>()
            .ok()
            .unwrap()
    }

    pub fn str() -> String {
        read()
    }

    pub fn s() -> Vec<char> {
        str().chars().collect()
    }

    pub fn i() -> i64 {
        read()
    }

    pub fn u() -> usize {
        read()
    }

    pub fn u2() -> (usize, usize) {
        (read(), read())
    }

    pub fn u3() -> (usize, usize, usize) {
        (read(), read(), read())
    }

    pub fn f() -> f64 {
        read()
    }

    pub fn c() -> char {
        read::<String>().pop().unwrap()
    }

    pub fn iv(n: usize) -> Vec<i64> {
        (0..n).map(|_| i()).collect()
    }

    pub fn iv2(n: usize) -> Vec<(i64, i64)> {
        (0..n).map(|_| iv(2)).map(|a| (a[0], a[1])).collect()
    }

    pub fn uv(n: usize) -> Vec<usize> {
        (0..n).map(|_| u()).collect()
    }

    pub fn uv2(n: usize) -> Vec<(usize, usize)> {
        (0..n).map(|_| uv(2)).map(|a| (a[0], a[1])).collect()
    }

    pub fn uv3(n: usize) -> Vec<(usize, usize, usize)> {
        (0..n).map(|_| uv(3)).map(|a| (a[0], a[1], a[2])).collect()
    }

    pub fn fv(n: usize) -> Vec<f64> {
        (0..n).map(|_| f()).collect()
    }

    pub fn cmap(h: usize) -> Vec<Vec<char>> {
        (0..h).map(|_| s()).collect()
    }
}