#include <cstdio>
#include <algorithm>
using namespace std;

typedef long long ll;

const int MAXN = 5000;
const ll INF = (1LL << 60);

ll dp[MAXN + 5];

int p[MAXN + 5], N, A, B;
int main() {
	scanf("%d%d%d", &N, &A, &B);
	for(int i=1;i<=N;i++) scanf("%d", &p[i]);
	p[N + 1] = N + 1;
	
	for(int i=1;i<=N+1;i++) {
		dp[i] = INF; int nw = 0; ll del = 0;
		for(int j=i-1;j>=1;j--) {
			if( p[j] < p[i] ) {
				if( nw < p[j] )
					dp[i] = min(dp[i], dp[j] + del), nw = p[j];
				del += B;
			}
			else del += A;
		}
		if( !nw ) dp[i] = del;
	}
	
	printf("%lld\n", dp[N + 1]);
}
### 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 <cstdio>
#include <algorithm>
using namespace std;

typedef long long ll;

const int MAXN = 5000;
const ll INF = (1LL << 60);

ll dp[MAXN + 5];

int p[MAXN + 5], N, A, B;
int main() {
	scanf("%d%d%d", &N, &A, &B);
	for(int i=1;i<=N;i++) scanf("%d", &p[i]);
	p[N + 1] = N + 1;
	
	for(int i=1;i<=N+1;i++) {
		dp[i] = INF; int nw = 0; ll del = 0;
		for(int j=i-1;j>=1;j--) {
			if( p[j] < p[i] ) {
				if( nw < p[j] )
					dp[i] = min(dp[i], dp[j] + del), nw = p[j];
				del += B;
			}
			else del += A;
		}
		if( !nw ) dp[i] = del;
	}
	
	printf("%lld\n", dp[N + 1]);
}
```