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)

/**
    *Makacha
**/
#include <bits/stdc++.h>
using namespace std;

template <typename T> inline void read (T &x) {bool b = 0; char c; while (!isdigit (c = getchar()) && c != '-');
if (c == '-') c = getchar(), b = 1; x = c - 48; while (isdigit(c = getchar())) x = (x<<3) + (x<<1) + c - 48; if (b)x=-x;}
template <typename T> inline void wrip(T x) {if (x > 9) wrip(x / 10); putchar(x%10 + 48); }

template <typename T> inline void write(T x) {if (x < 0) putchar('-'), x = -x; wrip(x); putchar(' ');}
template <typename T> inline void writeln(T x) {if (x < 0) putchar('-'), x = -x; wrip(x); putchar('\n');}
inline char readch() {char c; do c = getchar(); while (c == '\n' || c == ' '); return c;}
inline void reads(string &s) {char c; while((c=getchar())==' '||c =='\n'); s=c; while((c=getchar())!=' '&&c!='\n') s+=c;}
inline void getlines(string &s) {char c; while((c=getchar())=='\n'); s=c; while((c=getchar())!='\n') s+=c;}
typedef const int csint;

#define fi first
#define se second
#define mp make_pair
typedef long long ll;
typedef long double ld;
typedef pair <int, int> ii;
typedef vector <int> vi;

const ld pi = acos(-1);
const ll inf = 1e16;
const int mod = 1e9 + 7,
N = 5055;
int n, a[N];
ll A, B, f[N][N];
int main() {
    #ifdef makacha
        freopen("m.inp", "r", stdin);
        freopen("m.out", "w", stdout);
    #endif // makacha
    read(n);
    read(A);
    read(B);
    for (int i = 1; i <= n; i++)
        read(a[i]);
    for (int i = 0; i <= n; i++)
        for (int j = 0; j <= n; j++)
            f[i][j] = inf;
    f[1][a[1]] = 0;
    f[1][0] = A;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++) {
            f[i][j] = min(f[i][j], f[i][j - 1]);
            if (a[i + 1] >= j) {
                f[i + 1][a[i + 1]] = min(f[i + 1][a[i + 1]], f[i][j]);
                f[i + 1][j] = min(f[i + 1][j], f[i][j] + A);
            }
            else
                f[i + 1][j] = min(f[i + 1][j], f[i][j] + B);
        }
    cout << f[n][n];
}