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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <vector>
#include <algorithm>
#include <iostream>
using namespace std;
int main() {
    int n, a, b; cin >> n >> a >> b;
    int P[n]; for (int i = 0; i < n; i++) cin >> P[i];

    int lis[n] = {};
    fill(lis, lis+n, 1<<30);
    for (int i = 0; i < n; i++)
        *lower_bound(lis, lis+n, P[i]) = P[i];
    vector<int> s;
    int m = lower_bound(lis, lis+n, 1<<30) - lis - 1;
    // for (int i = 0; i <= m; i++) cerr << lis[i] << ' '; cerr << endl;
    int prevv = 1<<30;
    for (int i = n-1; i >= 0; i--) {
        if (lis[m] <= P[i] && P[i] < prevv) {
            s.push_back(i);
            prevv = P[i];
            m--;
        }
    }
    // for (int x : s) cerr << x << ' '; cerr << endl;
    reverse(s.begin(), s.end());
    bool in_lis[n] = {};
    for (int i : s) in_lis[i] = true;
    int acc_min[n], acc_max[n];
    int temp = 1<<30;
    for (int i = n-1; i >= 0; i--) {
        if (in_lis[i]) temp = min(temp, P[i]);
        acc_min[i] = temp;
    }
    temp = 0;
    for (int i = 0; i < n; i++) {
        if (in_lis[i]) temp = max(temp, P[i]);
        acc_max[i] = temp;
    }
    long long ans = 0;
    for (int i = 0; i < n; i++) {
        if (in_lis[i]) continue;
        auto it = lower_bound(s.begin(), s.end(), i);
        if (it != s.end() && P[i] > P[*it]) ans += a;
        else if (it != s.begin() && P[i] < P[*prev(it)]) ans += b;
    }
    cout << ans << endl;
}