#include<bits/stdc++.h>
using namespace std;
const int N = 5e3 + 11;
#define LL long long
LL f[N][N], A, B, a[N];
int n;
int main(){
	cin>>n>>A>>B;
	memset(f, 127, sizeof(f));
	f[0][0] = 0;
	for(int i = 1;i <= n; i++)cin>>a[i];
	for(int i = 0;i < n; i++){
		LL x = a[i+1];
		for(int j = 0;j <= n; j++){
			if(j <= x){
				f[i+1][j] = min(f[i+1][j], f[i][j] + A);
				f[i+1][x] = min(f[i+1][x], f[i][j]);
			}
			if(j > x){
				f[i+1][j] = min(f[i+1][j], f[i][j] + B);
			}
		}
	}
	LL ans = 1e18;
	for(int i = 0;i <= n; i++){
		ans = min(ans, f[n][i]);
	}
	cout<<ans<<endl;
	return 0;
}
### 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 = 5e3 + 11;
#define LL long long
LL f[N][N], A, B, a[N];
int n;
int main(){
	cin>>n>>A>>B;
	memset(f, 127, sizeof(f));
	f[0][0] = 0;
	for(int i = 1;i <= n; i++)cin>>a[i];
	for(int i = 0;i < n; i++){
		LL x = a[i+1];
		for(int j = 0;j <= n; j++){
			if(j <= x){
				f[i+1][j] = min(f[i+1][j], f[i][j] + A);
				f[i+1][x] = min(f[i+1][x], f[i][j]);
			}
			if(j > x){
				f[i+1][j] = min(f[i+1][j], f[i][j] + B);
			}
		}
	}
	LL ans = 1e18;
	for(int i = 0;i <= n; i++){
		ans = min(ans, f[n][i]);
	}
	cout<<ans<<endl;
	return 0;
}
```