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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll dp[5005][5005],a,b;
int p[5005],n;
int mx[5005][5005],mn[5005][5005];
int f[5005][5005];

int calc(int x1,int x2,int y1,int y2){
	return f[x2][y2]+f[x1-1][y1-1]-f[x1-1][y2]-f[x2][y1-1];
}

int main(){
	scanf("%d %lld %lld",&n,&a,&b);
	for(int i=1;i<=n;i++) scanf("%d",&p[i]),f[i][p[i]]=1;
	for(int l=1;l<=n;l++){
		mx[l][l]=mn[l][l]=p[l];
		for(int r=l+1;r<=n;r++){
			mx[l][r]=max(mx[l][r-1],p[r]);
			mn[l][r]=min(mn[l][r-1],p[r]);
		}
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++) f[i][j]=f[i][j-1]+f[i-1][j]-f[i-1][j-1]+f[i][j];
	}
	for(int len=2;len<=n;len++){
		for(int l=1,r;l<=n-len+1;l++){
			r=l+len-1;
			dp[l][r]=min(dp[l][r-1]+min(b*(p[r]<mx[l][r-1]),1ll*a*calc(l,r-1,p[r],n)),dp[l+1][r]+min(a*(p[l]>mn[l+1][r]),1ll*b*calc(l+1,r,1,p[r])));
		}
	}
	printf("%lld\n",dp[1][n]);

	return 0;
}