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: Accepted)

#include<bits/stdc++.h>
#define mod 1000000007
#define MOD(x) ((x)%mod)
#define pii pair<int,int>
#define mp(a,b) make_pair(a,b)
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
int n;
ll A,B;
int a[5005];
ll dp[2][5005];
int now,last;
int high[5005],low[5005];
int main()
{
	scanf("%d%lld%lld",&n,&A,&B);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
		for(int j=1;j<i;j++)
			if(a[j]>a[i]) high[i]++;
				else low[i]++;
	}
	for(int i=1;i<=n;i++)
	{
		last=now;
		now=!last;
		memset(dp[now],-1,sizeof dp[now]);
		dp[now][a[i]]=A*high[i]+B*low[i];
		for(int j=1;j<=n;j++)
		{
			if(dp[last][j]==-1) continue;
			if(a[i]>j) dp[now][j]=dp[last][j]+A;
			if(a[i]<j) dp[now][j]=dp[last][j]+B;
			if(a[i]>j) dp[now][a[i]]=min(dp[now][a[i]],dp[last][j]);
		}
	}
	ll ans=dp[now][1];
	for(int i=1;i<=n;i++)
		ans=min(ans,dp[now][i]);
	printf("%lld",ans);
	return 0;
}