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 N 320000
using namespace std;
long long n,A,B;
long long L[N],R[N];
long long f[5200][5200]; 
long long a[5200];
int main(){
	scanf("%lld%lld%lld",&n,&A,&B);
	if(n==1){
		printf("0\n");
		return 0;
	}
	for(long long i=1;i<=n;i++){
		long long x;
		scanf("%d",&x);
		a[x]=i;
	}
	for(long long i=1;i<=n;i++){
		for(long long j=i+1;j<=n;j++)if(a[j]<a[i]){L[i]=j;break;}
		for(long long j=i-1;j>=1;j--)if(a[j]>a[i]){R[i]=j;break;}
	}
	for(int i=0;i<=5100;i++){
		for(int j=0;j<=5100;j++)f[i][j]=1000000000000000LL;
	}
	f[0][n+1]=0;
	for(long long i=0;i<n;i++){
		for(long long j=n+1;j>i;j--){
			if(L[i+1]<j&&L[i+1]!=0)f[i+1][j]=min(f[i][j]+B,f[i+1][j]);
			else f[i+1][j]=min(f[i][j],f[i+1][j]);
			if(R[j-1]>i&&R[j-1]!=0)f[i][j-1]=min(f[i][j]+A,f[i][j-1]);
			else f[i][j-1]=min(f[i][j],f[i][j-1]);
		}
	}
	long long ans=1000000000000000LL;
	for(long long i=0;i<=n;i++)ans=min(ans,f[i][i+1]);
	printf("%lld\n",ans);
}