#include <bits/stdc++.h>

typedef long long ll;
const int N = 5054;

int n, L, R;
int p[N];
ll f[N][N];

inline void down(ll &x, const ll y) {x > y ? x = y : 0;}

int main() {
	int i, j;
	scanf("%d%d%d", &n, &R, &L);
	for (i = 1; i <= n; ++i) scanf("%d", p + i);
	for (i = 1; i <= n + 1; ++i) {
		for (j = 0; j < i; ++j) f[i][j] = f[i - 1][j] + (p[i] < p[j] ? L : R);
		f[i][i] = LLONG_MAX;
		for (j = 0; j < i; ++j) if (p[i] > p[j]) down(f[i][i], f[i - 1][j]);
	}
	printf("%lld\n", *std::min_element(f[n], f[n] + (n + 1)));;
	return 0;
}
### Prompt

Please provide a cpp coded solution 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>

typedef long long ll;
const int N = 5054;

int n, L, R;
int p[N];
ll f[N][N];

inline void down(ll &x, const ll y) {x > y ? x = y : 0;}

int main() {
	int i, j;
	scanf("%d%d%d", &n, &R, &L);
	for (i = 1; i <= n; ++i) scanf("%d", p + i);
	for (i = 1; i <= n + 1; ++i) {
		for (j = 0; j < i; ++j) f[i][j] = f[i - 1][j] + (p[i] < p[j] ? L : R);
		f[i][i] = LLONG_MAX;
		for (j = 0; j < i; ++j) if (p[i] > p[j]) down(f[i][i], f[i - 1][j]);
	}
	printf("%lld\n", *std::min_element(f[n], f[n] + (n + 1)));;
	return 0;
}
```