Problem:
Score : 1000 points

Problem Statement

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.

Sample Input 1

3 20 30
3 1 2

Sample Output 1

20

Shifting (p_1, p_2, p_3) to the left by one results in p = (1, 2, 3).

Sample Input 2

4 20 30
4 2 3 1

Sample Output 2

50

One possible sequence of operations is as follows:

Shift (p_1, p_2, p_3, p_4) to the left by one. Now we have p = (2, 3, 1, 4).

Shift (p_1, p_2, p_3) to the right by one. Now we have p = (1, 2, 3, 4).

Here, the total cost is 20 + 30 = 50.

Sample Input 3

1 10 10
1

Sample Output 3

0

Sample Input 4

4 1000000000 1000000000
4 3 2 1

Sample Output 4

3000000000

Sample Input 5

9 40 50
5 3 4 7 6 1 2 9 8

Sample Output 5

220

Solution in C++ (GCC 5.4.1) (judge verdict: Accepted)

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define N 5003
#define ll long long

using namespace std;

int n, m, a, b;
int pos[N];
ll f[N][N << 1];

int main(){
	cin >> n >> a >> b; m = n * 2 + 1;
	for(int i = 1; i <= n; ++i){
		int x; scanf("%d", &x);
		pos[x] = i * 2;
	}
	memset(f, 127, sizeof f);
	for(int j = 1; j <= m; ++j) f[0][j] = 0;
	for(int i = 1; i <= n; ++i)
		for(int j = 1; j <= m; ++j){
			ll tmp = (j != pos[i]) * (j < pos[i] ? b : a);
			tmp += f[i - 1][j - !(j & 1)]; 
			f[i][j] = min(f[i][j - 1], tmp);
		}
	cout << f[n][m] << endl;
	return 0;
}