#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int MN = 5005;
ll A, B, N, arr[MN], dp[MN][MN], i, j;
ll solve(ll n,ll lim){
    if(n==0) return 0;
    else if(dp[n][lim]!=-1) return dp[n][lim];
    if(arr[n]>lim) dp[n][lim]=A+solve(n-1,lim);
    else dp[n][lim]=min(solve(n-1,arr[n]),solve(n-1,lim)+B);
    return dp[n][lim];
}
int main(){
    for(scanf("%lld%lld%lld",&N,&A,&B),i=1;i<=N;i++)
        scanf("%lld",&arr[i]);
    memset(dp,-1,sizeof(dp));
    printf("%lld\n",solve(N,N+1));
    return 0;
}

### Prompt

Create a solution in CPP for 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;

typedef long long ll;
const int MN = 5005;
ll A, B, N, arr[MN], dp[MN][MN], i, j;
ll solve(ll n,ll lim){
    if(n==0) return 0;
    else if(dp[n][lim]!=-1) return dp[n][lim];
    if(arr[n]>lim) dp[n][lim]=A+solve(n-1,lim);
    else dp[n][lim]=min(solve(n-1,arr[n]),solve(n-1,lim)+B);
    return dp[n][lim];
}
int main(){
    for(scanf("%lld%lld%lld",&N,&A,&B),i=1;i<=N;i++)
        scanf("%lld",&arr[i]);
    memset(dp,-1,sizeof(dp));
    printf("%lld\n",solve(N,N+1));
    return 0;
}

```