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

#define MAX 5002

int n;
int A;
int B;

long long int dp[MAX][MAX];
bool us[MAX][MAX];


vector<int> v;


inline long long int dfs(int a, int b) {
	if (a == n) {
		return 0;
	}
	if (us[a][b])return dp[a][b];
	us[a][b] = true;
	dp[a][b] = LLONG_MAX;
	if (b == 0 || v[b - 1] < v[a]) {
		dp[a][b] = min(dp[a][b], dfs(a + 1, a + 1));
	}
	if (b == 0 || v[b - 1] < v[a]) {
		dp[a][b] = min(dp[a][b], dfs(a + 1, b) + A);
	}
	else {
		dp[a][b] = min(dp[a][b], dfs(a + 1, b) + B);
	}
	return dp[a][b];
}

int main() {
	cin >> n >> A >> B;
	for (int i = 0; i < n; i++) {
		int p;
		scanf("%d", &p);
		v.push_back(p);
	}
	printf("%lld\n", dfs(0, 0));
	return 0;
}

### 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"
using namespace std;

#define MAX 5002

int n;
int A;
int B;

long long int dp[MAX][MAX];
bool us[MAX][MAX];


vector<int> v;


inline long long int dfs(int a, int b) {
	if (a == n) {
		return 0;
	}
	if (us[a][b])return dp[a][b];
	us[a][b] = true;
	dp[a][b] = LLONG_MAX;
	if (b == 0 || v[b - 1] < v[a]) {
		dp[a][b] = min(dp[a][b], dfs(a + 1, a + 1));
	}
	if (b == 0 || v[b - 1] < v[a]) {
		dp[a][b] = min(dp[a][b], dfs(a + 1, b) + A);
	}
	else {
		dp[a][b] = min(dp[a][b], dfs(a + 1, b) + B);
	}
	return dp[a][b];
}

int main() {
	cin >> n >> A >> B;
	for (int i = 0; i < n; i++) {
		int p;
		scanf("%d", &p);
		v.push_back(p);
	}
	printf("%lld\n", dfs(0, 0));
	return 0;
}

```