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>
using namespace std;
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<n;i++)
#define all(v) v.begin(),v.end()
#define inf (int)(3e18)
#define P pair<int,int>
#define mod (int)(1e9+7)
template<class T> inline void chmin(T &a, T b) {
	a = min(a, b);
}
template<class T> inline void chmax(T &a, T b) {
	a = max(a, b);
}

int N,A,B;
int p[5005],q[5005];
int dp[5005][5005];
int lo[5005][5005],up[5005][5005];
signed main(){
	cin>>N>>A>>B;
	REP(i,N+1){
		cin>>p[i];
		q[p[i]]=i;
	}
	int cnt[5005];memset(cnt,0,sizeof(cnt));
	REP(i,N+1){
		for(int j=N;j>q[i];j--)lo[i][j]=cnt[j];
		for(int j=q[i];j>0;j--){
			cnt[j]++;lo[i][j]=cnt[j];
		}
	}
	memset(cnt,0,sizeof(cnt));
	for(int i=N;i>0;i--){
		rep(j,q[i])up[i][j]=cnt[j];
		for(int j=q[i];j<=N;j++){
			cnt[j]++;up[i][j]=cnt[j];
		}
	}
	rep(i,N+1)rep(j,N+2)dp[i][j]=inf;
	dp[0][N+1]=0;
	rep(i,N+1)for(int j=N+1;j>i+1;j--){
		if(i<N){
			if(q[i+1]+lo[i][q[i+1]+1]-up[j][q[i+1]-1]==i+1){
				chmin(dp[i+1][j],dp[i][j]);
			}else chmin(dp[i+1][j],dp[i][j]+B);
		}
		if(j>1){
			if(q[j-1]+lo[i][q[j-1]+1]-up[j][q[j-1]-1]==j-1){
				chmin(dp[i][j-1],dp[i][j]);
			}else chmin(dp[i][j-1],dp[i][j]+A);
		}
	}
	int ans=inf;
	rep(i,N+1){
		chmin(ans,dp[i][i+1]);
	}
	cout<<ans<<endl;
}