#include <bits/stdc++.h>

using namespace std;
const int N = 5e3 + 10;
typedef long long ll;

ll a , b , f[N];
int n , p[N];
int main(){
	scanf("%d%lld%lld",&n,&a,&b);
	for (int i = 1;i <= n;i++){
		scanf("%d",p + i);
	}
	p[n + 1] = n + 1;
	memset(f , 0x3f , sizeof f);
	f[0] = 0;
	
	for (int i = 1;i <= n + 1;i++){
		int g = 0;
		for (int j = i - 1;j >= 0;j--){
			if (p[j] > p[i]){
				g++;
			}else{
				f[i] = min(f[i] , f[j] + 1ll * g * a + 1ll * (i - j - 1 - g) * b);
			}
		}
	}
	cout<<f[n + 1];
	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 namespace std;
const int N = 5e3 + 10;
typedef long long ll;

ll a , b , f[N];
int n , p[N];
int main(){
	scanf("%d%lld%lld",&n,&a,&b);
	for (int i = 1;i <= n;i++){
		scanf("%d",p + i);
	}
	p[n + 1] = n + 1;
	memset(f , 0x3f , sizeof f);
	f[0] = 0;
	
	for (int i = 1;i <= n + 1;i++){
		int g = 0;
		for (int j = i - 1;j >= 0;j--){
			if (p[j] > p[i]){
				g++;
			}else{
				f[i] = min(f[i] , f[j] + 1ll * g * a + 1ll * (i - j - 1 - g) * b);
			}
		}
	}
	cout<<f[n + 1];
	return 0;
}
```