#include "bits/stdc++.h"
using namespace std;
int n;
int a[5005];
int A, B;
const long long inf = 1e16;
long long mem[5005][5005];

long long dp(int cur, int last) {
	if(cur == n + 1) return 0;
	long long &ans = mem[cur][last];
	if(ans != -1) return ans;
	ans = inf;
	if(last < a[cur]) {
		ans = min(ans, dp(cur + 1, a[cur]));
		ans = min(ans, dp(cur + 1, last) + A);
	} else {
		ans = min(ans, dp(cur + 1, last) + B);
	}
	return ans;
}

int main(int argc, char const *argv[])
{
	scanf("%d %d %d", &n, &A, &B);
	for(int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
	}
	memset(mem, -1, sizeof mem);
	printf("%lld\n", dp(1, 0));
	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;
int n;
int a[5005];
int A, B;
const long long inf = 1e16;
long long mem[5005][5005];

long long dp(int cur, int last) {
	if(cur == n + 1) return 0;
	long long &ans = mem[cur][last];
	if(ans != -1) return ans;
	ans = inf;
	if(last < a[cur]) {
		ans = min(ans, dp(cur + 1, a[cur]));
		ans = min(ans, dp(cur + 1, last) + A);
	} else {
		ans = min(ans, dp(cur + 1, last) + B);
	}
	return ans;
}

int main(int argc, char const *argv[])
{
	scanf("%d %d %d", &n, &A, &B);
	for(int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
	}
	memset(mem, -1, sizeof mem);
	printf("%lld\n", dp(1, 0));
	return 0;
}
```