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>
#include <assert.h>
using namespace std;
typedef long long ll;
typedef long double ld;
#define PB push_back
#define MP make_pair
#define MOD 1000000007LL
#define endl "\n"
#define fst first
#define snd second
const ll UNDEF = -1;
const ll INF=LLONG_MAX/4;
template<typename T> inline bool chkmax(T &aa, T bb) { return aa < bb ? aa = bb, true : false; }
template<typename T> inline bool chkmin(T &aa, T bb) { return aa > bb ? aa = bb, true : false; }
typedef pair<ll,ll> pll;
typedef vector<ll> vll;
typedef vector<int> vi;
int Rjump,Ljump;
const int mn=5004;
int a[mn];
// f(p,b) is the minimum cost such that:
// Position [1,p] with value within [1,b] are sorted
// Position [1,p] with value within [b+1,n] have jumped right
// f(p,b) = min f(p-1,[0,b]) + Ljump if a[p]<b + Rjump if a[p]>b
ll f[mn][mn],g[mn][mn];
int main(){	
	ios::sync_with_stdio(false);
	for (int x=0;x<mn;x++) for (int y=0;y<mn;y++) {
		f[x][y]=g[x][y]=INF;
	}
	int n; cin>>n>>Rjump>>Ljump;
	for (int p=1;p<=n;p++) {
		cin>>a[p];
	}
	for (int b=0;b<mn;b++) f[0][b]=g[0][b]=0;
	for (int p=1;p<=n;p++) {
		for (int b=1;b<=n;b++) {
			f[p][b]=g[p-1][b]+(a[p]<b?Ljump:0)+(a[p]>b?Rjump:0);
		}
		ll lof=f[p][1];
		for (int b=1;b<=n;b++) {
			chkmin(lof,f[p][b]);
			g[p][b]=lof;
		}
	}
	ll final=INF; for (int b=1;b<=n;b++) chkmin(final,f[n][b]);
	printf("%lld\n",final);
}