#include <bits/stdc++.h>

using namespace std;

const int N = 5005;
long long n, a, b, x[N];
long long f[N][N], pos[N], p[N][N];

int main() {
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	cin >> n >> a >> b;
	for(int i = 1; i <= n; ++i) cin >> x[i];
	memset(f, 127, sizeof(f));
	long long oo = f[0][0];
	f[0][0] = 0;
	for(int i = 1; i <= n; ++i) {
		pos[x[i]] = i;
	}
	for(int i = 0; i < n; ++i) {
		for(int j = 0; j <= n; ++j) {
			if(j < x[i + 1]) {
				f[i+1][j] = min(f[i+1][j], f[i][j] + a);
				f[i+1][x[i+1]] = min(f[i+1][x[i+1]], f[i][j]);
			}
			else{
				f[i+1][j] = min(f[i+1][j], f[i][j] + b);
			}
		}
	}
	for(int i = 0; i <= n; ++i) oo = min(oo, f[n][i]);
	cout << oo << '\n'; 
}
### Prompt

Your task is to create 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;

const int N = 5005;
long long n, a, b, x[N];
long long f[N][N], pos[N], p[N][N];

int main() {
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	cin >> n >> a >> b;
	for(int i = 1; i <= n; ++i) cin >> x[i];
	memset(f, 127, sizeof(f));
	long long oo = f[0][0];
	f[0][0] = 0;
	for(int i = 1; i <= n; ++i) {
		pos[x[i]] = i;
	}
	for(int i = 0; i < n; ++i) {
		for(int j = 0; j <= n; ++j) {
			if(j < x[i + 1]) {
				f[i+1][j] = min(f[i+1][j], f[i][j] + a);
				f[i+1][x[i+1]] = min(f[i+1][x[i+1]], f[i][j]);
			}
			else{
				f[i+1][j] = min(f[i+1][j], f[i][j] + b);
			}
		}
	}
	for(int i = 0; i <= n; ++i) oo = min(oo, f[n][i]);
	cout << oo << '\n'; 
}
```