#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 5e3 + 5;
int n,p[N],not_inv[N];
LL A,B,f[N][N];
//#define R
int main()
{
	cin>>n>>A>>B;
	for (int i = 1;i <= n; ++i)
		cin>>p[i]; 
	for (int i = 1;i <= n; ++i)
	{
		not_inv[i] = 0;
		for (int j = 1;j < i; ++j)
			if (p[j] < p[i])
				not_inv[i]++;
	}
	for (int i = 1;i <= n; ++i)
		for (int j = 1;j <= i; ++j)
			if (not_inv[i] + 1 > j)
				f[i][j] = f[i - 1][j];
			else
				f[i][j] = min(f[i - 1][j - 1] + B,f[i - 1][not_inv[i]] + A * ((LL)j - 1 - not_inv[i])); 
				//f[i - 1][not_inv[i]] + min(A,B * ((LL)j - 1 - not_inv[i]));
	#ifdef R
	for (int i = 1;i <= n; ++i)
	{
		for (int j = 1;j <= n; ++j)
			cout<<f[i][j]<<" ";
		puts("");
	}
	#endif
	cout<<f[n][n];
 } 
### Prompt

Generate 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<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 5e3 + 5;
int n,p[N],not_inv[N];
LL A,B,f[N][N];
//#define R
int main()
{
	cin>>n>>A>>B;
	for (int i = 1;i <= n; ++i)
		cin>>p[i]; 
	for (int i = 1;i <= n; ++i)
	{
		not_inv[i] = 0;
		for (int j = 1;j < i; ++j)
			if (p[j] < p[i])
				not_inv[i]++;
	}
	for (int i = 1;i <= n; ++i)
		for (int j = 1;j <= i; ++j)
			if (not_inv[i] + 1 > j)
				f[i][j] = f[i - 1][j];
			else
				f[i][j] = min(f[i - 1][j - 1] + B,f[i - 1][not_inv[i]] + A * ((LL)j - 1 - not_inv[i])); 
				//f[i - 1][not_inv[i]] + min(A,B * ((LL)j - 1 - not_inv[i]));
	#ifdef R
	for (int i = 1;i <= n; ++i)
	{
		for (int j = 1;j <= n; ++j)
			cout<<f[i][j]<<" ";
		puts("");
	}
	#endif
	cout<<f[n][n];
 } 
```