#include <bits/stdc++.h>

const int MAXN = 5010;
typedef long long LL;
int A[MAXN], va, vb, n;
LL dp[MAXN][MAXN];
void getmin(LL & x, const LL y) { x > y ? x = y : 0; }
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n >> va >> vb;
	for (int i = 1; i <= n; ++i) std::cin >> A[i];
	memset(dp, 0x3f, sizeof dp);
	**dp = 0;
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j <= n; ++j) {
			if (j > A[i]) {
				getmin(dp[i][j], dp[i - 1][j] + vb);
			} else {
				getmin(dp[i][A[i]], dp[i - 1][j]);
				getmin(dp[i][j], dp[i - 1][j] + va);
			}
		}
	}
	std::cout << *std::min_element(dp[n], dp[n] + n + 1) << std::endl;
	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>

const int MAXN = 5010;
typedef long long LL;
int A[MAXN], va, vb, n;
LL dp[MAXN][MAXN];
void getmin(LL & x, const LL y) { x > y ? x = y : 0; }
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n >> va >> vb;
	for (int i = 1; i <= n; ++i) std::cin >> A[i];
	memset(dp, 0x3f, sizeof dp);
	**dp = 0;
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j <= n; ++j) {
			if (j > A[i]) {
				getmin(dp[i][j], dp[i - 1][j] + vb);
			} else {
				getmin(dp[i][A[i]], dp[i - 1][j]);
				getmin(dp[i][j], dp[i - 1][j] + va);
			}
		}
	}
	std::cout << *std::min_element(dp[n], dp[n] + n + 1) << std::endl;
	return 0;
}

```