#include <bits/stdc++.h>
using namespace std;

const long long INF = 1e18;

int n, a, b;
int f[5001];
long long dp[5001][5001];

long long solve(int i, int prv) {
	if (i == n) return 0;
	if (dp[i][prv] != -1) return dp[i][prv];

	long long ans = INF;
	if (f[i] < prv) {
		ans = min(ans, solve(i+1, prv) + b);
	} else {
		ans = min(ans, solve(i+1, f[i]));
		ans = min(ans, solve(i+1, prv) + a);
	}
	return dp[i][prv] = ans;
}

signed main() {
	ios::sync_with_stdio(0); cin.tie(0);
	
	cin >> n >> a >> b;
	for (int i = 0; i < n; i++) {
		cin >> f[i];
	}
	memset(dp, -1, sizeof(dp));
	cout << solve(0, 0) << '\n';
 
	return 0;
}
### Prompt

Your challenge is to write 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;

const long long INF = 1e18;

int n, a, b;
int f[5001];
long long dp[5001][5001];

long long solve(int i, int prv) {
	if (i == n) return 0;
	if (dp[i][prv] != -1) return dp[i][prv];

	long long ans = INF;
	if (f[i] < prv) {
		ans = min(ans, solve(i+1, prv) + b);
	} else {
		ans = min(ans, solve(i+1, f[i]));
		ans = min(ans, solve(i+1, prv) + a);
	}
	return dp[i][prv] = ans;
}

signed main() {
	ios::sync_with_stdio(0); cin.tie(0);
	
	cin >> n >> a >> b;
	for (int i = 0; i < n; i++) {
		cin >> f[i];
	}
	memset(dp, -1, sizeof(dp));
	cout << solve(0, 0) << '\n';
 
	return 0;
}
```