#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
using namespace std;

int n,a,b,p[5005];
ll dp[5005][5005];

int main() {
	#ifdef kvutxdy
	freopen("C:/Users/khanh/OneDrive/RoadtoPurple/Code/Codeforces/input.in", "r", stdin);
	#endif
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	cin>>n>>a>>b;
	for(int i=1;i<=n;++i){
		int x;
		cin>>x;
		p[x]=i;
	}
	memset(dp,0x20,sizeof dp);
	dp[0][0]=0;
	for(int i=1;i<=n;++i)
		for(int j=0;j<i;++j)
			if(p[i]>p[j]){
				dp[i][i]=min(dp[i][i],dp[i-1][j]);
				dp[i][j]=min(dp[i][j],dp[i-1][j]+b);
			}
			else dp[i][j]=min(dp[i][j],dp[i-1][j]+a);
	cout<<*min_element(dp[n],dp[n]+n+1);
}
### 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>
#define fi first
#define se second
#define ll long long
using namespace std;

int n,a,b,p[5005];
ll dp[5005][5005];

int main() {
	#ifdef kvutxdy
	freopen("C:/Users/khanh/OneDrive/RoadtoPurple/Code/Codeforces/input.in", "r", stdin);
	#endif
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	cin>>n>>a>>b;
	for(int i=1;i<=n;++i){
		int x;
		cin>>x;
		p[x]=i;
	}
	memset(dp,0x20,sizeof dp);
	dp[0][0]=0;
	for(int i=1;i<=n;++i)
		for(int j=0;j<i;++j)
			if(p[i]>p[j]){
				dp[i][i]=min(dp[i][i],dp[i-1][j]);
				dp[i][j]=min(dp[i][j],dp[i-1][j]+b);
			}
			else dp[i][j]=min(dp[i][j],dp[i-1][j]+a);
	cout<<*min_element(dp[n],dp[n]+n+1);
}
```