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 <stdio.h>
#include <stdlib.h>
#define int long long
#define inf (int)(1e14)

int min(int a, int b){
	return a <= b ? a : b;
}

signed main(){
	int N, A, B, i, j;
	scanf("%lld%lld%lld", &N, &A, &B);
	int *p = (int *)malloc(sizeof(int) * (N + 1));
	int *p_inv = (int *)malloc(sizeof(int) * (N + 1));
	for(i = 1; i <= N; i++){
		scanf("%lld", &p[i]);
		p_inv[p[i]] = i;
	}
	int **dp1 = (int **)malloc(sizeof(int *) * (N + 1));
	int **dp2 = (int **)malloc(sizeof(int *) * (N + 1));
	int **m1 = (int **)malloc(sizeof(int *) * (N + 1));
	int **m2 = (int **)malloc(sizeof(int *) * (N + 1));
	for(i = 0; i <= N; i++){
		dp1[i] = (int *)malloc(sizeof(int) * (N + 1));
		dp2[i] = (int *)malloc(sizeof(int) * (N + 1));
		m1[i] = (int *)malloc(sizeof(int) * (N + 1));
		m2[i] = (int *)malloc(sizeof(int) * (N + 1));
	}
	for(j = 0; j <= N; j++){
		dp1[0][j] = 0;
		dp2[0][j] = 0;
		m1[0][j] = 0;
		m2[0][j] = 0;
	}
	for(i = 1; i <= N; i++){
		dp1[i][0] = inf;
		for(j = 1; j <= N; j++){
			dp1[i][j] = min(m1[i - 1][j - 1], m2[i - 1][j - 1]);
			if(j < p_inv[i]){
				dp1[i][j] += B;
			}
			if(j > p_inv[i]){
				dp1[i][j] += A;
			}
		}
		for(j = 0; j <= N; j++){
			dp2[i][j] = min(m1[i - 1][j], m2[i - 1][j]);
			if(j < p_inv[i]){
				dp2[i][j] += B;
			}
			else{
				dp2[i][j] += A;
			}
		}
		m1[i][0] = dp1[i][0];
		for(j = 1; j <= N; j++){
			m1[i][j] = min(m1[i][j - 1], dp1[i][j]);
		}
		m2[i][0] = dp2[i][0];
		for(j = 1; j <= N; j++){
			m2[i][j] = min(m2[i][j - 1], dp2[i][j]);
		}
	}
	printf("%lld\n", min(m1[N][N], m2[N][N]));
	return 0;
}