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)

use std::io::Read;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).ok();
    let mut iter = buf.split_whitespace();
    let n = iter.next().unwrap().parse::<usize>().unwrap();
    let k = iter.next().unwrap().parse::<i64>().unwrap();
    let a = (0..n).map(|_| iter.next().unwrap().parse::<i32>().unwrap()).collect::<Vec<_>>();
    let mut res : i64 = 0;
    let m = 1000000007;
    let mul =  (k * (k-1) / 2) % m;
    for i in 0..n {
        let mut cnt0 = 0;
        let mut cnt1 = 0;
        for j in 0..n {
            if a[j] < a[i] {
                cnt0 += 1;
                if i < j { cnt1 += 1; }
            }
        }
        res = (res + k * cnt1) % m;
        res = (res + mul * cnt0) % m;
    }
    println!("{}", res);
}