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

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N; ll A,B; cin >> N >> A >> B;
    vector<int> p(N);
    for (int i=0;i<N;++i) cin >> p[i];
    vector<vector<ll>> dp(N+1,vector<ll>(N+1,INF));
    dp[0][0]=0;
    for (int i=0;i<N;++i){
        for (int j=0;j<=N;++j){
            if (p[i]>j){
                dp[i+1][p[i]]=min(dp[i+1][p[i]],dp[i][j]);
                dp[i+1][j]=min(dp[i+1][j],dp[i][j]+A);
            } else dp[i+1][j]=min(dp[i+1][j],dp[i][j]+B);
        }
    }
    ll ans=INF;
    for (int j=0;j<=N;++j) ans=min(ans,dp[N][j]);
    cout << ans << '\n';
}
### 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;
typedef long long ll;
const long long INF=1e18;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N; ll A,B; cin >> N >> A >> B;
    vector<int> p(N);
    for (int i=0;i<N;++i) cin >> p[i];
    vector<vector<ll>> dp(N+1,vector<ll>(N+1,INF));
    dp[0][0]=0;
    for (int i=0;i<N;++i){
        for (int j=0;j<=N;++j){
            if (p[i]>j){
                dp[i+1][p[i]]=min(dp[i+1][p[i]],dp[i][j]);
                dp[i+1][j]=min(dp[i+1][j],dp[i][j]+A);
            } else dp[i+1][j]=min(dp[i+1][j],dp[i][j]+B);
        }
    }
    ll ans=INF;
    for (int j=0;j<=N;++j) ans=min(ans,dp[N][j]);
    cout << ans << '\n';
}
```