#include <bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < (int)(n); i ++)
#define rep1(i, n) for(int i = 1; i <= (int)(n); i ++)
#define MP make_pair

using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const LL INF = 1e16 + 7;

int n, a[5005];
LL A, B;
LL dp[5005];

int main()
{
	scanf("%d%lld%lld", &n, &A, &B);
	rep1(i, n) scanf("%d", &a[i]);
	a[n + 1] = n + 1;
	n ++;
	
	dp[0] = 0;
	rep1(i, n) dp[i] = INF;
	
	rep1(i, n) {
		int cntr = 0, cmin = a[i];
		for(int j = i - 1; j >= 0; j --) {
			if(a[j] > a[i]) cntr ++;
			else {
				if(cmin > a[i]) continue;
				int cntl = i - j - 1 - cntr;
				dp[i] = min(dp[i], dp[j] + A * cntr + B * cntl);
				cmin = a[i];
			}
		}
	}
	printf("%lld\n", dp[n]);
	return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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>
#define rep(i, n) for(int i = 0; i < (int)(n); i ++)
#define rep1(i, n) for(int i = 1; i <= (int)(n); i ++)
#define MP make_pair

using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const LL INF = 1e16 + 7;

int n, a[5005];
LL A, B;
LL dp[5005];

int main()
{
	scanf("%d%lld%lld", &n, &A, &B);
	rep1(i, n) scanf("%d", &a[i]);
	a[n + 1] = n + 1;
	n ++;
	
	dp[0] = 0;
	rep1(i, n) dp[i] = INF;
	
	rep1(i, n) {
		int cntr = 0, cmin = a[i];
		for(int j = i - 1; j >= 0; j --) {
			if(a[j] > a[i]) cntr ++;
			else {
				if(cmin > a[i]) continue;
				int cntl = i - j - 1 - cntr;
				dp[i] = min(dp[i], dp[j] + A * cntr + B * cntl);
				cmin = a[i];
			}
		}
	}
	printf("%lld\n", dp[n]);
	return 0;
}
```