#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

int num[5005],pos[5005],cnt[5005];
ll f[5005][5005];

int main() {
  memset(f,0x3f,sizeof(f));
  int n,a,b;
  scanf("%d%d%d",&n,&a,&b);
  for(int i=1;i<=n;i++) {
  	scanf("%d",&num[i]);
  	pos[num[i]]=i;
  }
  for(int i=0;i<=n;i++) f[i][0]=f[0][i]=0;
  for(int i=1;i<=n;i++) {
  	for(int j=num[i]+1;j<=n;j++)
  	  if (pos[j]<i) cnt[j]++;
    for(int j=1;j<=n;j++)
      if (pos[j]<=i) {
      	if (!cnt[j]) f[i][j]=f[i][j-1];
      	else f[i][j]=min(f[i][j-1]+a,f[pos[j]-1][j]+(ll)b*cnt[j]);
	  }
      else f[i][j]=f[i][j-1];
  }
  printf("%lld\n",f[n][n]);
  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>

using namespace std;

typedef long long ll;

int num[5005],pos[5005],cnt[5005];
ll f[5005][5005];

int main() {
  memset(f,0x3f,sizeof(f));
  int n,a,b;
  scanf("%d%d%d",&n,&a,&b);
  for(int i=1;i<=n;i++) {
  	scanf("%d",&num[i]);
  	pos[num[i]]=i;
  }
  for(int i=0;i<=n;i++) f[i][0]=f[0][i]=0;
  for(int i=1;i<=n;i++) {
  	for(int j=num[i]+1;j<=n;j++)
  	  if (pos[j]<i) cnt[j]++;
    for(int j=1;j<=n;j++)
      if (pos[j]<=i) {
      	if (!cnt[j]) f[i][j]=f[i][j-1];
      	else f[i][j]=min(f[i][j-1]+a,f[pos[j]-1][j]+(ll)b*cnt[j]);
	  }
      else f[i][j]=f[i][j-1];
  }
  printf("%lld\n",f[n][n]);
  return 0;
}
```