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>
using namespace std;
#define int long long
#define in ({int x=0;int c=getchar(),n=0;for(;!isdigit(c);c=getchar()) n=(c=='-');for(;isdigit(c);c=getchar()) x=x*10+c-'0';n?-x:x;})
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int l,int r){return l+rng()%(r-l+1);}
#define fasty ios_base::sync_with_stdio(0),cin.tie(0);
#define forinc(a,b,c) for(int a=b,_c=c;a<=_c;++a)
#define fordec(a,b,c) for(int a=b,_c=c;a>=_c;--a)
#define forv(a,b) for(auto&a:b)
#define fi first
#define se second
#define pb push_back
#define ii pair<int,int>
#define mt make_tuple
#define all(a) a.begin(),a.end()
#define reset(f, x) memset(f, x, sizeof(f))
#define bit(x,i) ((x>>(i-1))&1)
#define on(x,i) (x|(1ll<<(i-1)))
#define off(x,i) (x&~(1<<(i-1)))
#define gg exit(0);

const int N=5010;

int n,a,b;
int p[N],f[N];

main(){
    #define task "Rotation"
    //freopen(task".inp","r",stdin);
    //freopen(task".out","w",stdout);

    n=in,a=in,b=in;
    forinc(i,1,n) p[i]=in;
    p[n+1]=n+1, n++;
    reset(f,127);
    f[0]=0;
    forinc(i,1,n){
        int cntx=0,cnty=0;
        fordec(j,i-1,0){
            if(p[j]<p[i]) f[i]=min(f[i],f[j]+cntx*b+cnty*a);
            if(p[j]<p[i]) cntx++; else cnty++;
        }
    }
    cout<<f[n];
}