#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int MAXN = 5e3 + 5;
const ll INF = 1e18;

int N, A, B;
int perm[MAXN], inv[MAXN];
ll dp[MAXN][MAXN];

void load() {
	scanf("%d%d%d", &N, &A, &B);
	for (int i = 1; i <= N; i++) {
		scanf("%d", perm + i);
		inv[perm[i]] = i;
	}
}

ll rec(int num, int mx) {
	ll &ref = dp[num][mx];
	if (ref != INF)
		return ref;
	if (num > N)
		return ref = 0;
	if (mx > inv[num]) 
		return ref = rec(num + 1, mx) + A;
	return ref = min(rec(num + 1, inv[num]), rec(num + 1, mx) + B);
}

ll solve() {
	for (int i = 0; i < MAXN; i++)
		for (int j = 0; j < MAXN; j++)
			dp[i][j] = INF;
			
	return rec(1, 0);
}

int main() {
	load();
	printf("%lld\n", solve());
	return 0;
}
### Prompt

Create a solution in CPP for 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;

typedef long long ll;

const int MAXN = 5e3 + 5;
const ll INF = 1e18;

int N, A, B;
int perm[MAXN], inv[MAXN];
ll dp[MAXN][MAXN];

void load() {
	scanf("%d%d%d", &N, &A, &B);
	for (int i = 1; i <= N; i++) {
		scanf("%d", perm + i);
		inv[perm[i]] = i;
	}
}

ll rec(int num, int mx) {
	ll &ref = dp[num][mx];
	if (ref != INF)
		return ref;
	if (num > N)
		return ref = 0;
	if (mx > inv[num]) 
		return ref = rec(num + 1, mx) + A;
	return ref = min(rec(num + 1, inv[num]), rec(num + 1, mx) + B);
}

ll solve() {
	for (int i = 0; i < MAXN; i++)
		for (int j = 0; j < MAXN; j++)
			dp[i][j] = INF;
			
	return rec(1, 0);
}

int main() {
	load();
	printf("%lld\n", solve());
	return 0;
}
```