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++ (GCC 9.2.1) (judge verdict: Accepted)

// IOI 2021
#include <bits/stdc++.h>
using namespace std;

#define endl '\n'
#define ends ' '
#define die(x) return cout << x << endl, 0
#define all(v) v.begin(), v.end()
#define sz(x) (int)(x.size())
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) { cerr << ends << H; debug_out(T...); }
#define debug(...) cerr << "{" << #__VA_ARGS__ << "}:", debug_out(__VA_ARGS__)
typedef long long ll;
typedef pair<int, int> pii;
const int INF = 1e9;
const ll MOD = 1e9 + 7;
 
////////////////////////////////////////////////////////////////////

const int N = 5e3 + 5;

int P[N], C[N][N], M[N][N];
ll DP[N][N];
// C[i][j] : # k : i <= k <= j && A[i] > A[k] 
// M[i][j] : index maximum A[k] <= j && k <= i
// DP[i][j] : answer for first i element and numbers less than j
int main() {

	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	mt19937 Rnd(time(0));

	int n, A, B; cin >> n >> A >> B;
	for (int i = 1; i <= n; i++) cin >> P[i];
	for (int i = 1; i <= n; i++) for (int j = i + 1; j <= n; j++) C[i][j] = C[i][j - 1] + (P[i] > P[j]);
	for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) M[i][j] = (P[i] <= j && P[i] > P[M[i - 1][j]]) ?  i : M[i - 1][j];
	for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) {
		if (!M[i][j]) continue;
		DP[i][j] = min(A + DP[i][P[M[i][j]] - 1], DP[M[i][j] - 1][j] + 1LL * C[M[i][j]][i] * B);
	}
	cout << DP[n][n] << endl;

	return 0;
}