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(non_snake_case)]
#![allow(unused_variables)]

fn main() {
    let (N, K): (usize, i64) = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        let mut iter = line.split_whitespace();
        (
            iter.next().unwrap().parse().unwrap(),
            iter.next().unwrap().parse().unwrap()
        )
    };
    let A: Vec<i64> = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.split_whitespace()
            .map(|x| x.parse().unwrap())
            .collect()
    };

    let MOD = 1_000_000_007;
    let ans = {
        let mut res = 0;
        for i in 0..N {
            let mut x = 0;
            for j in (i + 1)..N {
                if A[i] > A[j] {
                    x += 1;
                }
            }
            let mut y = 0;
            for j in 0..i {
                if A[i] > A[j] {
                    y += 1;
                }
            }
            res = (res + x * (K * (K + 1)) / 2) % MOD;
            res = (res + y * (K * (K - 1)) / 2) % MOD;
        }
        res
    };

    println!("{}", ans);
}