#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
#define chmax(a,b) a=max(a,b);
#define chmin(a,b) a=min(a,b);
#define N 5010
ll f[N][N],n,a,b,p[N];
int main(){
    cin>>n>>a>>b;
    for(int i=0;i<n;i++)cin>>p[i];
    for(int i=0;i<N;i++)for(int j=0;j<N;j++)f[i][j]=1e17;
    f[0][0]=0;
    for(int i=0;i<n;i++){
	for(int j=0;j<=n;j++){
	    if(j<p[i]){
		chmin(f[i+1][p[i]],f[i][j]);
		chmin(f[i+1][j],f[i][j]+a);
	    }
	    else chmin(f[i+1][j],f[i][j]+b);
	}
    }
    ll ans=1e17;
    for(int i=0;i<=n;i++)chmin(ans,f[n][i]);
    cout<<ans<<endl;
    return 0;
}


### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
typedef long long ll;
typedef pair<ll,ll> P;
#define chmax(a,b) a=max(a,b);
#define chmin(a,b) a=min(a,b);
#define N 5010
ll f[N][N],n,a,b,p[N];
int main(){
    cin>>n>>a>>b;
    for(int i=0;i<n;i++)cin>>p[i];
    for(int i=0;i<N;i++)for(int j=0;j<N;j++)f[i][j]=1e17;
    f[0][0]=0;
    for(int i=0;i<n;i++){
	for(int j=0;j<=n;j++){
	    if(j<p[i]){
		chmin(f[i+1][p[i]],f[i][j]);
		chmin(f[i+1][j],f[i][j]+a);
	    }
	    else chmin(f[i+1][j],f[i][j]+b);
	}
    }
    ll ans=1e17;
    for(int i=0;i<=n;i++)chmin(ans,f[n][i]);
    cout<<ans<<endl;
    return 0;
}


```