#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL N=5010;
LL n, a, b, p[N], id[N], dp[N][N], s[N][N];
int main(){
	scanf("%lld%lld%lld",&n,&a,&b);
	for(LL i=1; i<=n; ++i)scanf("%lld",&p[i]), id[p[i]]=i;
	memset(s,0x3f,sizeof(s));
	for(LL i=0; i<=n; ++i)s[0][i]=0;
	for(LL i=1; i<=n; ++i)
		for(LL j=0; j<=n; ++j){
			dp[i][j]=s[i-1][j]+(j>=id[i]?a:b);
			if(j==id[i])dp[i][j]=min(dp[i][j],s[i-1][j-1]);
			if(j>0)s[i][j]=min(s[i][j-1],dp[i][j]);else s[i][j]=dp[i][j];
		}
	LL mn=0x3f3f3f3f3f3f3f3f;
	for(LL i=1; i<=n; ++i)mn=min(mn,dp[n][i]);
	printf("%lld",mn);
	return 0;
}
### Prompt

Create a solution in cpp for 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 LL N=5010;
LL n, a, b, p[N], id[N], dp[N][N], s[N][N];
int main(){
	scanf("%lld%lld%lld",&n,&a,&b);
	for(LL i=1; i<=n; ++i)scanf("%lld",&p[i]), id[p[i]]=i;
	memset(s,0x3f,sizeof(s));
	for(LL i=0; i<=n; ++i)s[0][i]=0;
	for(LL i=1; i<=n; ++i)
		for(LL j=0; j<=n; ++j){
			dp[i][j]=s[i-1][j]+(j>=id[i]?a:b);
			if(j==id[i])dp[i][j]=min(dp[i][j],s[i-1][j-1]);
			if(j>0)s[i][j]=min(s[i][j-1],dp[i][j]);else s[i][j]=dp[i][j];
		}
	LL mn=0x3f3f3f3f3f3f3f3f;
	for(LL i=1; i<=n; ++i)mn=min(mn,dp[n][i]);
	printf("%lld",mn);
	return 0;
}
```