#include<bits/stdc++.h>
using namespace std;
const long long Inf=1e18;

int gi() {
	int x=0,o=1;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
	if(ch=='-') o=-1,ch=getchar();
	while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
	return x*o;
}

int main() {
	int n=gi(),a=gi(),b=gi();
	vector<int> p(n+1);
	for(int i=1;i<=n;i++) p[i]=gi();
	vector<int> q(n+1);
	for(int i=1;i<=n;i++) q[p[i]]=i;
	vector<long long> f(1,0);
	for(int i=1;i<=n;i++) {
		vector<long long> new_f(i+1,Inf);
		for(int j=0;j<i;j++) {
			if(q[i]>q[j]) {
				new_f[i]=min(new_f[i],f[j]);
				new_f[j]=min(new_f[j],f[j]+b);
			}
			else {
				new_f[j]=min(new_f[j],f[j]+a);
			}
		}
		swap(f,new_f);
	}
	printf("%lld\n",*min_element(f.begin(),f.end()));
	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;
const long long Inf=1e18;

int gi() {
	int x=0,o=1;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
	if(ch=='-') o=-1,ch=getchar();
	while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
	return x*o;
}

int main() {
	int n=gi(),a=gi(),b=gi();
	vector<int> p(n+1);
	for(int i=1;i<=n;i++) p[i]=gi();
	vector<int> q(n+1);
	for(int i=1;i<=n;i++) q[p[i]]=i;
	vector<long long> f(1,0);
	for(int i=1;i<=n;i++) {
		vector<long long> new_f(i+1,Inf);
		for(int j=0;j<i;j++) {
			if(q[i]>q[j]) {
				new_f[i]=min(new_f[i],f[j]);
				new_f[j]=min(new_f[j],f[j]+b);
			}
			else {
				new_f[j]=min(new_f[j],f[j]+a);
			}
		}
		swap(f,new_f);
	}
	printf("%lld\n",*min_element(f.begin(),f.end()));
	return 0;
}

```