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 <iostream>
#include <string>
#include <array>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <algorithm>
#include <cmath>
#include <iomanip>

#define FOR(i, a, b) for(decltype(b) i = (a); i < (b); ++i)
#define REP(i, n) FOR(i, 0, n)

typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;

using namespace std;

#ifdef INPUT
#include <sstream>
stringstream input(R"(
9 40 50
5 3 4 7 6 1 2 9 8
)");
#define cin input
#endif

int main()
{
	ll n, a, b;
	cin >> n >> a >> b;

	vector<int> p(n);
	for (auto& i : p)
		cin >> i;

	auto eval = [&](int v, bool s)
	{
		ll r = 0;
		int x = 1;
		//cout << '_';
		for (int i = 0; i < n; ++i)
		{
			if (x == v + 1)
				break;
			if (p[i] > v)
			{
				if (!s)
				{
					r += (v - x + 1) * b;
					//cout << "++" << (v - x + 1);
					break;
				}
				continue;
			}
			if (p[i] > x)
			{
				r += (p[i] - x) * b;
				//cout << '+' << (p[i] - x);
			}
			if (x <= p[i])
				x = p[i] + 1;
		}
		x = n;
		for (int i = n - 1; i >= 0; --i)
		{
			if (x == v)
				break;
			if (p[i] <= v)
			{
				if (s)
				{
					r += (x - v) * a;
					//cout << "--" << (x - v);
					break;
				}
				continue;
			}
			if (p[i] < x)
			{
				r += (x - p[i]) * a;
				//cout << '-' << (x - p[i]);
			}
			if (x >= p[i])
				x = p[i] - 1;
		}
		//cout << '>' << v << s << ' ' << r << endl;
		return r;
	};

	ll r = n * (a + b);	
	REP (i, n + 1)
		r = min(r, min(eval(i, true), eval(i, false)));

	cout << r << endl;

	return 0;
}