#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

const int N=5e3+5;
int n,R,L,a[N];ll ans=1e18,dp[N];

int main() {
	scanf("%d%d%d",&n,&R,&L);
	memset(dp,64,sizeof(dp)),dp[0]=0;
	for(int i=1;i<=n;++i) scanf("%d",&a[i]);
	for(int i=1;i<=n;++i) {
		for(int j=0;j<a[i];++j) dp[a[i]]=min(dp[a[i]],dp[j]);
		for(int j=0;j<a[i];++j) dp[j]+=R;
		for(int j=a[i];j<n;++j) dp[j+1]+=L;
	}
	for(int i=0;i<=n;++i) ans=min(ans,dp[i]);
	printf("%lld\n",ans);
	return 0;
}
### Prompt

Your challenge is to write 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 <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

const int N=5e3+5;
int n,R,L,a[N];ll ans=1e18,dp[N];

int main() {
	scanf("%d%d%d",&n,&R,&L);
	memset(dp,64,sizeof(dp)),dp[0]=0;
	for(int i=1;i<=n;++i) scanf("%d",&a[i]);
	for(int i=1;i<=n;++i) {
		for(int j=0;j<a[i];++j) dp[a[i]]=min(dp[a[i]],dp[j]);
		for(int j=0;j<a[i];++j) dp[j]+=R;
		for(int j=a[i];j<n;++j) dp[j+1]+=L;
	}
	for(int i=0;i<=n;++i) ans=min(ans,dp[i]);
	printf("%lld\n",ans);
	return 0;
}
```