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)

#![allow(unused_imports)]
#![allow(non_snake_case)]
use std::io::*;
use std::str::FromStr;
 
pub fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed reading char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed parsing")
}
 
use std::cmp::{min, max};
use std::num;
use std::collections::HashMap;
use std::collections::HashSet;

const MOD : u64 = 1_000_000_000 + 7;

fn main() {
    
    let n = read::<usize>();
    let k = read::<u64>();
    
    let a : Vec<u64> = (0..n).map(|_| read::<u64>() ).collect();

    let mut p = 0;
    let mut q = 0;
    
    for i in 0..n-1 {
        for j in i..n {
            if a[i] > a[j] {
                p += 1;
            }
        }
    }
    
    for i in 0..n {
        for j in 0..n {
            if a[i] > a[j] {
                q += 1;
            }
        }
    }
    
    let ans = ((p * k) % MOD + q * ((k * (k-1) / 2) % MOD) % MOD) % MOD;
    
    
    println!("{}",ans);
}