Problem:
Score : 1000 points

Problem Statement

You are given a permutation p = (p_1, \ldots, p_N) of \{ 1, \ldots, N \}.
You can perform the following two kinds of operations repeatedly in any order:

Pay a cost A. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the left by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_{l + 1}, p_{l + 2}, \ldots, p_r, p_l, respectively.

Pay a cost B. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the right by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_r, p_l, \ldots, p_{r - 2}, p_{r - 1}, respectively.

Find the minimum total cost required to sort p in ascending order.

Constraints

All values in input are integers.

1 \leq N \leq 5000

1 \leq A, B \leq 10^9

(p_1 \ldots, p_N) is a permutation of \{ 1, \ldots, N \}.

Input

Input is given from Standard Input in the following format:

N A B
p_1 \cdots p_N

Output

Print the minimum total cost required to sort p in ascending order.

Sample Input 1

3 20 30
3 1 2

Sample Output 1

20

Shifting (p_1, p_2, p_3) to the left by one results in p = (1, 2, 3).

Sample Input 2

4 20 30
4 2 3 1

Sample Output 2

50

One possible sequence of operations is as follows:

Shift (p_1, p_2, p_3, p_4) to the left by one. Now we have p = (2, 3, 1, 4).

Shift (p_1, p_2, p_3) to the right by one. Now we have p = (1, 2, 3, 4).

Here, the total cost is 20 + 30 = 50.

Sample Input 3

1 10 10
1

Sample Output 3

0

Sample Input 4

4 1000000000 1000000000
4 3 2 1

Sample Output 4

3000000000

Sample Input 5

9 40 50
5 3 4 7 6 1 2 9 8

Sample Output 5

220

Solution in Rust (1.42.0) (judge verdict: Accepted)

// ---------- begin chmin, chmax ----------
trait ChangeMinMax {
    fn chmin(&mut self, x: Self) -> bool;
    fn chmax(&mut self, x: Self) -> bool;
}

impl<T: PartialOrd> ChangeMinMax for T {
    fn chmin(&mut self, x: Self) -> bool {
        *self > x && {
            *self = x;
            true
        }
    }
    fn chmax(&mut self, x: Self) -> bool {
        *self < x && {
            *self = x;
            true
        }
    }
}
// ---------- end chmin, chmax ----------

use proconio::*;

fn run() {
    input! {
        n: usize,
        a: u64,
        b: u64,
        p: [usize; n],
    }
    let inf = 10u64.pow(9) * 5000 + 1;
    let mut dp = vec![inf; n + 1];
    dp[0] = 0;
    for p in p {
        let mut next = vec![inf; n + 1];
        next[p].chmin(*dp[..=p].iter().min().unwrap());
        for (next, dp) in next.iter_mut().zip(dp.iter()).take(p) {
            next.chmin(*dp + a);
        }
        for (next, dp) in next.iter_mut().zip(dp.iter()).skip(p + 1) {
            next.chmin(*dp + b);
        }
        dp = next;
    }
    let ans = dp.into_iter().min().unwrap();
    println!("{}", ans);
}

fn main() {
    run();
}