#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
typedef long long ll;
const int NC = 5e3 + 5;
int N, A, B, p[NC];
ll dp[NC][NC];
int main()
{
	scanf("%d%d%d", &N, &A, &B);
	for (int i = 1; i <= N; ++i) scanf("%d", &p[i]);
	memset(dp, 0x3f, sizeof(dp));
	dp[0][0] = 0;
	for (int i = 1; i <= N; ++i)
		for (int j = 0; j <= N; ++j)
		{
			if (p[i] > j)
			{
				dp[i][j] = min(dp[i][j], dp[i-1][j] + A);
				dp[i][p[i]] = min(dp[i][p[i]], dp[i-1][j]);
			}
			else dp[i][j] = min(dp[i][j], dp[i-1][j] + B);
		}
	ll ans = 0x3f3f3f3f3f3f3f3f;
	for (int i = 1; i <= N; ++i) ans = min(ans, dp[N][i]);
	printf("%lld\n", ans);
	return 0;
}
### Prompt

Generate 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 <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
typedef long long ll;
const int NC = 5e3 + 5;
int N, A, B, p[NC];
ll dp[NC][NC];
int main()
{
	scanf("%d%d%d", &N, &A, &B);
	for (int i = 1; i <= N; ++i) scanf("%d", &p[i]);
	memset(dp, 0x3f, sizeof(dp));
	dp[0][0] = 0;
	for (int i = 1; i <= N; ++i)
		for (int j = 0; j <= N; ++j)
		{
			if (p[i] > j)
			{
				dp[i][j] = min(dp[i][j], dp[i-1][j] + A);
				dp[i][p[i]] = min(dp[i][p[i]], dp[i-1][j]);
			}
			else dp[i][j] = min(dp[i][j], dp[i-1][j] + B);
		}
	ll ans = 0x3f3f3f3f3f3f3f3f;
	for (int i = 1; i <= N; ++i) ans = min(ans, dp[N][i]);
	printf("%lld\n", ans);
	return 0;
}
```