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 ri register
#define int long long
using namespace std; const int N=5010;
inline int read()
{
    int s=0,w=1; char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-') w=-1; ch=getchar(); }
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+(ch^48),ch=getchar(); return s*w;
}
int n,A,B,val[N];
int dp[N][N],p[N];
signed main()
{
	n=read(), A=read(), B=read();
	for(ri int i=1;i<=n;i++) {int x=read(); p[x]=i; }
	for(ri int i=1;i<=n;i++) for(ri int j=0;j<=n;j++) dp[i][j]=LONG_LONG_MAX/2;
	for(ri int i=1;i<=n;i++)
	{
		for(ri int j=0;j<=n;j++)
		{
			if(p[i]==j)
			{
				dp[i][j]=min(dp[i-1][j-1],dp[i][j]);
				continue;
			}
			dp[i][j]=min(dp[i][j],dp[i-1][j]+(j<p[i]?B:A));
		}
		for(ri int j=1;j<=n;j++) dp[i][j]=min(dp[i][j],dp[i][j-1]);
	}
	printf("%lld\n",dp[n][n]);
    return 0;
}