#include <bits/stdc++.h>
typedef long long ll;
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
using namespace std;
const ll MAX_N=5000,INF=1e18;
ll N,A,B,memo[MAX_N+1][MAX_N*2+1],q[MAX_N];
ll dfs(ll n,ll x){
	if(n==N){
		return 0;
	}
	if(memo[n][x]!=-1){
		return memo[n][x];
	}
	ll ret=INF,x2;
	if(q[n]*2+1>x){
		x2=(x+1)/2*2;
		ret=dfs(n+1,x2)+B;
		ret=min(ret,dfs(n+1,q[n]*2+1));
	}
	ret=min(ret,A+dfs(n+1,max(q[n]*2+2,(x+1)/2*2)));
	return memo[n][x]=ret;
}
	
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>N>>A>>B;
	REP(i,N){
		ll p;
		cin>>p;
		p--;
		q[p]=i;
	}
	memset(memo,-1,sizeof(memo));
	cout<<dfs(0,0)<<endl;
}
### 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>
typedef long long ll;
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
using namespace std;
const ll MAX_N=5000,INF=1e18;
ll N,A,B,memo[MAX_N+1][MAX_N*2+1],q[MAX_N];
ll dfs(ll n,ll x){
	if(n==N){
		return 0;
	}
	if(memo[n][x]!=-1){
		return memo[n][x];
	}
	ll ret=INF,x2;
	if(q[n]*2+1>x){
		x2=(x+1)/2*2;
		ret=dfs(n+1,x2)+B;
		ret=min(ret,dfs(n+1,q[n]*2+1));
	}
	ret=min(ret,A+dfs(n+1,max(q[n]*2+2,(x+1)/2*2)));
	return memo[n][x]=ret;
}
	
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>N>>A>>B;
	REP(i,N){
		ll p;
		cin>>p;
		p--;
		q[p]=i;
	}
	memset(memo,-1,sizeof(memo));
	cout<<dfs(0,0)<<endl;
}
```