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

const int N = 5010;

typedef long long LL;

int num[N]; LL dp[N];

int main() {
    int n, A, B; scanf("%d%d%d", &n, &A, &B);
    for (int i = 1; i <= n; i++) scanf("%d", &num[i]);
    memset(dp, 0x3f, sizeof(dp)), dp[0] = 0, num[n + 1] = n + 1, n++;
    for (int i = 1; i <= n; i++) {
        for (int j = i - 1, l = 0, r = 0, mx = 0; j >= 0; j--) {
            if (num[j] < num[i] && num[j] > mx) mx = num[j];
            if (mx <= num[j] && num[j] < num[i]) 
                dp[i] = min(dp[i], dp[j] + (LL)l * B + (LL)r * A);
            if (num[j] < num[i]) l++;
            else r++;
        }
    }
    printf("%lld\n", dp[n]);
}
### Prompt

Generate a CPP solution to 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;

const int N = 5010;

typedef long long LL;

int num[N]; LL dp[N];

int main() {
    int n, A, B; scanf("%d%d%d", &n, &A, &B);
    for (int i = 1; i <= n; i++) scanf("%d", &num[i]);
    memset(dp, 0x3f, sizeof(dp)), dp[0] = 0, num[n + 1] = n + 1, n++;
    for (int i = 1; i <= n; i++) {
        for (int j = i - 1, l = 0, r = 0, mx = 0; j >= 0; j--) {
            if (num[j] < num[i] && num[j] > mx) mx = num[j];
            if (mx <= num[j] && num[j] < num[i]) 
                dp[i] = min(dp[i], dp[j] + (LL)l * B + (LL)r * A);
            if (num[j] < num[i]) l++;
            else r++;
        }
    }
    printf("%lld\n", dp[n]);
}
```