#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 5010;
int f[N], a[N], n, va, vb, mn[N][N];
signed main() {
    scanf("%lld%lld%lld", &n, &va, &vb);
    for (int i = 1; i <= n; ++i) scanf("%lld", a + i);
    if (va > vb) {
        swap(va, vb);
        reverse(a + 1, a + n + 1);
        for (int i = 1; i <= n; ++i) a[i] = n - a[i] + 1;
    }
    ++n; a[n] = n;
    for (int i = 1; i <= n; ++i)
        for (int j = i + 1; j <= n; ++j)
            mn[i][j] = mn[i][j - 1] + bool(a[j] < a[i]);
    for (int i = 1; i <= n; ++i) {
        f[i] = 1e15;
        for (int j = i - 1; ~j; --j) if (a[j] < a[i])
            f[i] = min(f[i], f[j] + mn[j][i] * vb + (i - j - 1 - mn[j][i]) * va);
    }
    printf("%lld\n", f[n]);
    return 0;
}
### 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;
#define int long long
const int N = 5010;
int f[N], a[N], n, va, vb, mn[N][N];
signed main() {
    scanf("%lld%lld%lld", &n, &va, &vb);
    for (int i = 1; i <= n; ++i) scanf("%lld", a + i);
    if (va > vb) {
        swap(va, vb);
        reverse(a + 1, a + n + 1);
        for (int i = 1; i <= n; ++i) a[i] = n - a[i] + 1;
    }
    ++n; a[n] = n;
    for (int i = 1; i <= n; ++i)
        for (int j = i + 1; j <= n; ++j)
            mn[i][j] = mn[i][j - 1] + bool(a[j] < a[i]);
    for (int i = 1; i <= n; ++i) {
        f[i] = 1e15;
        for (int j = i - 1; ~j; --j) if (a[j] < a[i])
            f[i] = min(f[i], f[j] + mn[j][i] * vb + (i - j - 1 - mn[j][i]) * va);
    }
    printf("%lld\n", f[n]);
    return 0;
}
```