#include<bits/stdc++.h>
const int maxn = 5001;
typedef long long ll;
int p[maxn], n;
ll dp[maxn], a, b;
int main()
{
	std::cin >> n >> a >> b;
	for(int i = 1;i <= n;++i) std::cin >> p[i], dp[i] = (i - 1) * a;
	ll ans = 1e18;
	for(int i = 1;i <= n;++i) {
		ll s = 0; int min = 1e9;
		for(int j = i + 1;j <= n;++j) {
			if(p[i] < p[j] && p[j] < min) {
				min = p[j];
				dp[j] = std::min(dp[j], dp[i] + s);
			}
			s += p[j] < p[i] ? b : a;
		}
		ans = std::min(ans, dp[i] + (n - i) * b);
	}
	std::cout << ans << '\n';
}

### 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>
const int maxn = 5001;
typedef long long ll;
int p[maxn], n;
ll dp[maxn], a, b;
int main()
{
	std::cin >> n >> a >> b;
	for(int i = 1;i <= n;++i) std::cin >> p[i], dp[i] = (i - 1) * a;
	ll ans = 1e18;
	for(int i = 1;i <= n;++i) {
		ll s = 0; int min = 1e9;
		for(int j = i + 1;j <= n;++j) {
			if(p[i] < p[j] && p[j] < min) {
				min = p[j];
				dp[j] = std::min(dp[j], dp[i] + s);
			}
			s += p[j] < p[i] ? b : a;
		}
		ans = std::min(ans, dp[i] + (n - i) * b);
	}
	std::cout << ans << '\n';
}

```