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

typedef long long ll;
const int maxn = 5e3 + 5;

int C[maxn];
ll dp[maxn][maxn], A, B;
int n, where[maxn], a[maxn];

int main(int argc, char const *argv[])
{
	// freopen("in", "r", stdin);
	cin >> n >> A >> B;
	for(int i = 1; i <= n; i++) {
		cin >> a[i];
		where[a[i]] = i;
	}
	for(int i = 1; i <= n; i++) {
		int c = 0;
		for(int j = 1; j < i; j++) {
			if(where[j] < where[i])
				c++;
		}
		C[i] = c + 1;
	}
	for(int i = 0; i <= n; i++) dp[i][0] = 0;
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= i; j++) {
			if(C[i] > j) {
				dp[i][j] = dp[i - 1][j];
			} else {
				dp[i][j] = min(dp[i - 1][j - 1] + A, dp[i - 1][C[i] - 1] + (j - C[i]) * B);
			}
		}
	}
	cout << dp[n][n] << endl;
	return 0;
}
### Prompt

Please formulate 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;

typedef long long ll;
const int maxn = 5e3 + 5;

int C[maxn];
ll dp[maxn][maxn], A, B;
int n, where[maxn], a[maxn];

int main(int argc, char const *argv[])
{
	// freopen("in", "r", stdin);
	cin >> n >> A >> B;
	for(int i = 1; i <= n; i++) {
		cin >> a[i];
		where[a[i]] = i;
	}
	for(int i = 1; i <= n; i++) {
		int c = 0;
		for(int j = 1; j < i; j++) {
			if(where[j] < where[i])
				c++;
		}
		C[i] = c + 1;
	}
	for(int i = 0; i <= n; i++) dp[i][0] = 0;
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= i; j++) {
			if(C[i] > j) {
				dp[i][j] = dp[i - 1][j];
			} else {
				dp[i][j] = min(dp[i - 1][j - 1] + A, dp[i - 1][C[i] - 1] + (j - C[i]) * B);
			}
		}
	}
	cout << dp[n][n] << endl;
	return 0;
}
```