#include <bits/stdc++.h>
using i64 = long long;

int main() {
    int n; i64 a, b;
    std::cin >> n >> a >> b;
    std::vector<int> p(n + 1);
    for (int i = 1; i <= n; i++) {
        int in;
        std::cin >> in;
        p[in] = i * 2 - 1;
    }
    std::vector<std::vector<i64>> dp(n + 1, std::vector<i64>(2 * n + 2));
    std::vector<i64> v(2 * n + 2);
    for (int i = 1; i <= n; i++) {
        v[0] = dp[i - 1][0];
        for (int j = 0; j < 2 * n + 1; j++) {
            v[j + 1] = std::min(v[j], dp[i - 1][j + 1]);
        }
        for (int j = 0; j <= 2 * n + 1; j++) {
            dp[i][j] = v[j] + (p[i] < j ? a : p[i] == j ? 0 : b);
        }
    }
    std::cout << *std::min_element(dp[n].begin(), dp[n].end()) << std::endl;

    return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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.

Examples

Input

3 20 30
3 1 2


Output

20


Input

4 20 30
4 2 3 1


Output

50


Input

1 10 10
1


Output

0


Input

4 1000000000 1000000000
4 3 2 1


Output

3000000000


Input

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


Output

220

### Response

```cpp
#include <bits/stdc++.h>
using i64 = long long;

int main() {
    int n; i64 a, b;
    std::cin >> n >> a >> b;
    std::vector<int> p(n + 1);
    for (int i = 1; i <= n; i++) {
        int in;
        std::cin >> in;
        p[in] = i * 2 - 1;
    }
    std::vector<std::vector<i64>> dp(n + 1, std::vector<i64>(2 * n + 2));
    std::vector<i64> v(2 * n + 2);
    for (int i = 1; i <= n; i++) {
        v[0] = dp[i - 1][0];
        for (int j = 0; j < 2 * n + 1; j++) {
            v[j + 1] = std::min(v[j], dp[i - 1][j + 1]);
        }
        for (int j = 0; j <= 2 * n + 1; j++) {
            dp[i][j] = v[j] + (p[i] < j ? a : p[i] == j ? 0 : b);
        }
    }
    std::cout << *std::min_element(dp[n].begin(), dp[n].end()) << std::endl;

    return 0;
}

```