#include <bits/stdc++.h>

typedef long long LL;
const int N = 5005;
const LL inf = 0x3f3f3f3f3f3f3f3f;

int n, a, b;
LL dp[N][N], ans = inf;

void down(LL &x, LL y) { x = std::min(x, y); }
int main() {
	std::ios::sync_with_stdio(0), std::cin.tie(0);
	std::memset(dp, 0x3f, sizeof dp), std::cin >> n >> a >> b;
	dp[0][0] = 0;
	for (int i = 0; i < n; ++i) {
		int x; std::cin >> x;
		for (int j = 0; j <= n; ++j) if (dp[i][j] < inf) {
			if (x < j)
				down(dp[i + 1][j], dp[i][j] + b);
			else
				down(dp[i + 1][x], dp[i][j]),
				down(dp[i + 1][j], dp[i][j] + a);
		}
	}
	for (int i = 0; i <= n; ++i) down(ans, dp[n][i]);
	std::cout << ans << '\n';
	return 0;
}
### Prompt

Please create a solution in CPP 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>

typedef long long LL;
const int N = 5005;
const LL inf = 0x3f3f3f3f3f3f3f3f;

int n, a, b;
LL dp[N][N], ans = inf;

void down(LL &x, LL y) { x = std::min(x, y); }
int main() {
	std::ios::sync_with_stdio(0), std::cin.tie(0);
	std::memset(dp, 0x3f, sizeof dp), std::cin >> n >> a >> b;
	dp[0][0] = 0;
	for (int i = 0; i < n; ++i) {
		int x; std::cin >> x;
		for (int j = 0; j <= n; ++j) if (dp[i][j] < inf) {
			if (x < j)
				down(dp[i + 1][j], dp[i][j] + b);
			else
				down(dp[i + 1][x], dp[i][j]),
				down(dp[i + 1][j], dp[i][j] + a);
		}
	}
	for (int i = 0; i <= n; ++i) down(ans, dp[n][i]);
	std::cout << ans << '\n';
	return 0;
}
```