#include<bits/stdc++.h>
#define ll long long
#define INF 1e18
#define N 5010
using namespace std;
template <typename T> void read(T &x){
	x=0;char c=getchar();int fh=1;
	while (!isdigit(c)){if (c=='-')fh=-1;c=getchar();}
	while (isdigit(c))x=x*10+c-'0',c=getchar();
	x*=fh;
}
ll n,yo,zu,mi[N],a[N],dp[N][N];
int main(){
	read(n);read(yo);read(zu);mi[0]=INF;
	for (int i=1;i<=n;i++) read(a[i]);
	for (int i=1;i<=n+1;i++){
		for (int j=1;j<=n;j++) mi[j]=min(mi[j-1],dp[i-1][j]);
		for (int j=1;j<=n;j++){
			dp[i][j]=mi[j];
			if (j<a[i])dp[i][j]+=yo;
			if (j>a[i])dp[i][j]+=zu;
		}
	}
	cout<<mi[n]<<endl;
	return 0;
}
### Prompt

Develop a solution in CPP 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 ll long long
#define INF 1e18
#define N 5010
using namespace std;
template <typename T> void read(T &x){
	x=0;char c=getchar();int fh=1;
	while (!isdigit(c)){if (c=='-')fh=-1;c=getchar();}
	while (isdigit(c))x=x*10+c-'0',c=getchar();
	x*=fh;
}
ll n,yo,zu,mi[N],a[N],dp[N][N];
int main(){
	read(n);read(yo);read(zu);mi[0]=INF;
	for (int i=1;i<=n;i++) read(a[i]);
	for (int i=1;i<=n+1;i++){
		for (int j=1;j<=n;j++) mi[j]=min(mi[j-1],dp[i-1][j]);
		for (int j=1;j<=n;j++){
			dp[i][j]=mi[j];
			if (j<a[i])dp[i][j]+=yo;
			if (j>a[i])dp[i][j]+=zu;
		}
	}
	cout<<mi[n]<<endl;
	return 0;
}
```