#include <stdio.h>
#include <algorithm>
using namespace std;
const int maxn = 5050;
long long f[2 * maxn], nf[2 * maxn];
int p[maxn];
int main(){
	int n, A, B; scanf("%d %d %d", &n, &A, &B);
	for(int i = 1; i <= n; i++){
		int x; scanf("%d", &x);
		p[x] = 2 * i;
	}
	int l = 2 * n + 1;
	for(int i = 1; i <= n; i++){
		long long mi = f[1];
		f[1] += B;
		for(int j = 2; j <= l; j++){
			mi = min(mi, f[j]);
			long long cost = (p[i] < j) * A + (p[i] > j) * B;
			f[j] = mi + cost;
		}
	}
	long long ans = 1ll<<62;
	for(int j = 1; j <= l; j++) ans = min(ans, f[j]);
	printf("%lld\n", ans);
	return 0;
}

### Prompt

In CPP, your task is to solve 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 <stdio.h>
#include <algorithm>
using namespace std;
const int maxn = 5050;
long long f[2 * maxn], nf[2 * maxn];
int p[maxn];
int main(){
	int n, A, B; scanf("%d %d %d", &n, &A, &B);
	for(int i = 1; i <= n; i++){
		int x; scanf("%d", &x);
		p[x] = 2 * i;
	}
	int l = 2 * n + 1;
	for(int i = 1; i <= n; i++){
		long long mi = f[1];
		f[1] += B;
		for(int j = 2; j <= l; j++){
			mi = min(mi, f[j]);
			long long cost = (p[i] < j) * A + (p[i] > j) * B;
			f[j] = mi + cost;
		}
	}
	long long ans = 1ll<<62;
	for(int j = 1; j <= l; j++) ans = min(ans, f[j]);
	printf("%lld\n", ans);
	return 0;
}

```