#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=5005;
ll a[N],dp[N][N];
inline void cmn(ll&a,ll b){if(a>b)a=b;}
int main()
{
	ll n,A,B,ans=1ll<<60;scanf("%lld%lld%lld",&n,&A,&B);
	for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
	memset(dp,0x3f,sizeof(dp));dp[0][0]=0;
	for(int i=1;i<=n;i++)for(int j=0;j<=n;j++)if(a[i]>j)cmn(dp[i][j],dp[i-1][j]+A),cmn(dp[i][a[i]],dp[i-1][j]);else cmn(dp[i][j],dp[i-1][j]+B);
	for(int i=0;i<=n;i++)cmn(ans,dp[n][i]);
	printf("%lld\n",ans);
	return 0;
}
### Prompt

In Cpp, your task is to solve 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;
#define ll long long
const int N=5005;
ll a[N],dp[N][N];
inline void cmn(ll&a,ll b){if(a>b)a=b;}
int main()
{
	ll n,A,B,ans=1ll<<60;scanf("%lld%lld%lld",&n,&A,&B);
	for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
	memset(dp,0x3f,sizeof(dp));dp[0][0]=0;
	for(int i=1;i<=n;i++)for(int j=0;j<=n;j++)if(a[i]>j)cmn(dp[i][j],dp[i-1][j]+A),cmn(dp[i][a[i]],dp[i-1][j]);else cmn(dp[i][j],dp[i-1][j]+B);
	for(int i=0;i<=n;i++)cmn(ans,dp[n][i]);
	printf("%lld\n",ans);
	return 0;
}
```