#include <bits/stdc++.h>
using namespace std;
#define modulo 1000000007
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 10000000000000002




int main(){

	int N;
	cin>>N;
	
	long long A,B;
	cin>>A>>B;
	
	vector<int> P(N);
	for(int i=0;i<N;i++)cin>>P[i];
	P.insert(P.begin(),0);
	P.push_back(N+1);
	
	vector<long long> dp(P.size(),Inf);
	dp[0] = 0LL;
	
	for(int i=1;i<P.size();i++){
		long long t = 0LL;
		for(int j=i-1;j>=0;j--){
			if(P[j]>P[i]){
				t += A;
			}
			else{
				dp[i] = min(dp[i],dp[j]+t);
				t += B;
			}
		}
	}
	cout<<dp.back()<<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;
#define modulo 1000000007
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 10000000000000002




int main(){

	int N;
	cin>>N;
	
	long long A,B;
	cin>>A>>B;
	
	vector<int> P(N);
	for(int i=0;i<N;i++)cin>>P[i];
	P.insert(P.begin(),0);
	P.push_back(N+1);
	
	vector<long long> dp(P.size(),Inf);
	dp[0] = 0LL;
	
	for(int i=1;i<P.size();i++){
		long long t = 0LL;
		for(int j=i-1;j>=0;j--){
			if(P[j]>P[i]){
				t += A;
			}
			else{
				dp[i] = min(dp[i],dp[j]+t);
				t += B;
			}
		}
	}
	cout<<dp.back()<<endl;
	return 0;
}
```