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)

#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

#define F first
#define S second
#define int long long
#define ll long long
//#define int  unsigned long long
#define pb push_back
//#define double long double

using namespace std;
using namespace __gnu_pbds;
typedef  tree< int , null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
//const int mod = 998244353;
const int N = 7005;
const int K = 1000000;
vector < int > ls[N], rs[N];
int dp[N][N], p[N][N], c[N];
main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    //freopen("input.txt", "r", stdin);
    int b, a, n;
    cin >> n >> a >> b;
    for (int i = 1; i <= n; i++){
        for (int j = 1; j <= n; j++){
            if (i == 0 && j == 0) continue;
            dp[i][j] = 1e16;
        }
    }
    for (int i = 1; i <= n; i++) cin >> c[i];
    for (int i = 1; i <= n; i++){
        for (int j = 0; j <= n; j++){
            if (c[i] == j){
                dp[i][j] = p[i - 1][j];
            } else{
                if (i < c[i]){
                    dp[i][j] = dp[i - 1][j] + a;
                }
                if (i > c[i]){
                    dp[i][j] = dp[i - 1][j] + b;
                }
                if (i == c[i]){
                    dp[i][j] = dp[i - 1][j] + min(a, b);
                }
            }
            if (j != 0)
                p[i][j] = min(p[i][j - 1], dp[i][j]); else
                p[i][j] = dp[i][j];
        }
    }
    int ans = 1e18;
    for (int i = 0; i <= n; i++){
        ans = min(ans, dp[n][i]);
    }
    cout << ans;
}