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

using ll = long long;
const ll INF = 1e18;

const int MAXN = 5010;
int N;
ll A, B;
int P[MAXN];
ll dp[MAXN];

int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> N >> A >> B;
	for (int i = 1; i <= N; i++) {
		cin >> P[i];
	}
	P[0] = 0;
	P[N+1] = N+1;

	dp[0] = 0;
	for (int i = 1; i <= N+1; i++) {
		dp[i] = INF;
		int numBigger = 0;
		for (int j = i-1; j >= 0; j--) {
			if (P[j] < P[i]) {
				dp[i] = min(dp[i], dp[j] + numBigger * A + (i-j-1-numBigger) * B);
			} else {
				numBigger++;
			}
		}
	}
	cout << dp[N+1] << '\n';

	return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;

using ll = long long;
const ll INF = 1e18;

const int MAXN = 5010;
int N;
ll A, B;
int P[MAXN];
ll dp[MAXN];

int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> N >> A >> B;
	for (int i = 1; i <= N; i++) {
		cin >> P[i];
	}
	P[0] = 0;
	P[N+1] = N+1;

	dp[0] = 0;
	for (int i = 1; i <= N+1; i++) {
		dp[i] = INF;
		int numBigger = 0;
		for (int j = i-1; j >= 0; j--) {
			if (P[j] < P[i]) {
				dp[i] = min(dp[i], dp[j] + numBigger * A + (i-j-1-numBigger) * B);
			} else {
				numBigger++;
			}
		}
	}
	cout << dp[N+1] << '\n';

	return 0;
}

```