#include<bits/stdc++.h>
using namespace std;

int main() {
    int n, a, b;
    cin >> n >> a >> b;
    vector<long long> dp(n + 1, 1e18);
    dp[0] = 0;
    vector<int> p(n);
    for (int i = 0; i < n; ++i) {
        cin >> p[i];
        dp[i + 1] = dp[0];
        dp[0] += a;
        for (int j = 0; j < i; ++j) {
            if (p[j] < p[i]) dp[i + 1] = min(dp[i + 1], dp[j + 1]);
            dp[j + 1] += p[j] < p[i] ? a : b;
        }
    }
    long long ans = 1e18;
    for (int i = 0; i <= n; ++i) ans = min(ans, dp[i]);
    cout << ans << '\n';
}
### Prompt

In Cpp, your task is to solve the following problem:

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 namespace std;

int main() {
    int n, a, b;
    cin >> n >> a >> b;
    vector<long long> dp(n + 1, 1e18);
    dp[0] = 0;
    vector<int> p(n);
    for (int i = 0; i < n; ++i) {
        cin >> p[i];
        dp[i + 1] = dp[0];
        dp[0] += a;
        for (int j = 0; j < i; ++j) {
            if (p[j] < p[i]) dp[i + 1] = min(dp[i + 1], dp[j + 1]);
            dp[j + 1] += p[j] < p[i] ? a : b;
        }
    }
    long long ans = 1e18;
    for (int i = 0; i <= n; ++i) ans = min(ans, dp[i]);
    cout << ans << '\n';
}
```