#include<bits/stdc++.h>
using namespace std;
 
#define ll long long
#define pb push_back
 
const int maxn = 5e3 + 20;

ll dp[maxn] , a[maxn];

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	int n;
	ll x , y;
	cin >> n >> x >> y;

	for(int i = 1; i <= n; i++)
		cin >> a[i];

	a[n + 1] = n + 1;
	memset(dp , 63 , sizeof dp);
	dp[0] = 0;
	for(int i = 1; i <= n + 1; i++)
	{
		int mx = -1 , t = 0;
		for(int j = i - 1; j >= 0; j--)
		{
			if(a[j] > a[i])
				t++;
			else if(a[j] > mx)
			{
				mx = a[j];
				dp[i] = min(dp[i] , dp[j] + t * x + (i - j - 1 - t) * y);
			}
		}
	}

	cout << dp[n + 1] << endl;
}

















### 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;
 
#define ll long long
#define pb push_back
 
const int maxn = 5e3 + 20;

ll dp[maxn] , a[maxn];

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	int n;
	ll x , y;
	cin >> n >> x >> y;

	for(int i = 1; i <= n; i++)
		cin >> a[i];

	a[n + 1] = n + 1;
	memset(dp , 63 , sizeof dp);
	dp[0] = 0;
	for(int i = 1; i <= n + 1; i++)
	{
		int mx = -1 , t = 0;
		for(int j = i - 1; j >= 0; j--)
		{
			if(a[j] > a[i])
				t++;
			else if(a[j] > mx)
			{
				mx = a[j];
				dp[i] = min(dp[i] , dp[j] + t * x + (i - j - 1 - t) * y);
			}
		}
	}

	cout << dp[n + 1] << endl;
}

















```