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 ll long long
#define ull unsigned long long
#define db double
#define pb push_back
#define ppb pop_back
#define F first
#define S second
#define mp make_pair
#define all(x) (x).begin(), (x).end()

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;

const int N = 5003;
int n, a, b, p[N], ind[N], fen[N];
ll dp[N];

void upd(int x, int y) {
	for (; x <= n + 1; x |= (x + 1)) {
		fen[x] += y;
	}
}

int get(int x) {
	int res = 0;
	for (; x >= 0; x = (x & (x + 1)) - 1) {
		res += fen[x];
	}
	return res;
}

int main() {
	ios_base::sync_with_stdio(false), cin.tie(NULL);
	#ifdef LOCAL
		freopen("input.txt", "r", stdin);
	#endif
	
	cin >> n >> a >> b;
	p[0] = 0;
	p[n + 1] = n + 1;
	for (int i = 1; i <= n; i++) {
		cin >> p[i];
		ind[p[i]] = i;
	}
	ind[0] = 0;
	ind[n + 1] = n + 1;
	memset(& dp, 0x3f, sizeof(dp));
	dp[0] = 0;
	for (int i = 1; i <= n + 1; i++) {
		for (int j = p[i] - 1; j >= 0; j--) {
			if (ind[j] < i) {
				ll inner = get(i - 1) - get(ind[j]);
				ll l = get(ind[j]);
				ll r = get(n + 1) - get(i);
				dp[i] = min(dp[i], dp[ind[j]] + inner * min(a, b) + l * a + r * b);
			}
			upd(ind[j], 1);
		}
		for (int j = p[i] - 1; j >= 0; j--) {
			upd(ind[j], -1);
		}
	}
	cout << dp[n + 1];
}