#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define f(i, x, n) for (int i = x; i < (int)(n); ++i)

int const N = 5000;
ll dp[N + 1][N + 1];
int x[N + 1], in[N + 1];

inline void up(ll &x, ll y) { if (y < x)x = y; }

int main(){
	int n, a, b;
	scanf("%d%d%d", &n, &a, &b);
	f(i, 1, n + 1)scanf("%d", x + i);
	f(i, 1, n + 1)in[x[i]] = i;
	f(i, 0, n + 1)f(j, 0, n + 1)dp[i][j] = 1e18;
	dp[0][0] = 0;
	f(p, 0, n + 1)f(i, 0, n){
		int lc = in[i + 1];
		up(dp[p][i + 1], dp[p][i] + (lc < p ? a : b));
		if (lc >= p)up(dp[lc][i + 1], dp[p][i]);
	}
	ll an = 1e18;
	f(i, 0, n + 1)up(an, dp[i][n]);
	printf("%lld\n", an);
}
### 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>
using namespace std;
#define ll long long
#define f(i, x, n) for (int i = x; i < (int)(n); ++i)

int const N = 5000;
ll dp[N + 1][N + 1];
int x[N + 1], in[N + 1];

inline void up(ll &x, ll y) { if (y < x)x = y; }

int main(){
	int n, a, b;
	scanf("%d%d%d", &n, &a, &b);
	f(i, 1, n + 1)scanf("%d", x + i);
	f(i, 1, n + 1)in[x[i]] = i;
	f(i, 0, n + 1)f(j, 0, n + 1)dp[i][j] = 1e18;
	dp[0][0] = 0;
	f(p, 0, n + 1)f(i, 0, n){
		int lc = in[i + 1];
		up(dp[p][i + 1], dp[p][i] + (lc < p ? a : b));
		if (lc >= p)up(dp[lc][i + 1], dp[p][i]);
	}
	ll an = 1e18;
	f(i, 0, n + 1)up(an, dp[i][n]);
	printf("%lld\n", an);
}
```