#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
#define ll long long
const int Q=5005;
ll f[Q][Q];
int p[Q];
int main()
{
	int n;
	ll a,b;
	scanf("%d%lld%lld",&n,&a,&b);
	for(int i=1;i<=n;i++)
		scanf("%d",&p[i]);
	for(int i=1;i<=n;i++){
        f[i][p[i]]=998244353;
        for(int j=p[i];j;--j)
            f[i][j]=f[i-1][j]+a,f[i][p[i]]=min(f[i][p[i]],f[i-1][j]);
        for(int j=p[i]+1;j<=n;j++)
            f[i][j]=f[i-1][j]+b;
    }
    ll als=100000000000000000LL;
    for(int j=1;j<=n;j++)
        als=min(als,f[n][j]);
    printf("%lld\n",als);
	return 0;
}

### Prompt

Please formulate a cpp solution to 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 <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
#define ll long long
const int Q=5005;
ll f[Q][Q];
int p[Q];
int main()
{
	int n;
	ll a,b;
	scanf("%d%lld%lld",&n,&a,&b);
	for(int i=1;i<=n;i++)
		scanf("%d",&p[i]);
	for(int i=1;i<=n;i++){
        f[i][p[i]]=998244353;
        for(int j=p[i];j;--j)
            f[i][j]=f[i-1][j]+a,f[i][p[i]]=min(f[i][p[i]],f[i-1][j]);
        for(int j=p[i]+1;j<=n;j++)
            f[i][j]=f[i-1][j]+b;
    }
    ll als=100000000000000000LL;
    for(int j=1;j<=n;j++)
        als=min(als,f[n][j]);
    printf("%lld\n",als);
	return 0;
}

```