#include <bits/stdc++.h>
#define f first
#define s second
using namespace std;
using ll = long long;
using ii = pair<int, int>;
constexpr int MN = 5005;

ll p[MN], DP[MN][MN];

int main() {
    if (fopen("in", "r")) freopen("in", "r", stdin), freopen("out", "w", stdout);
    ios_base::sync_with_stdio(0), cin.tie(0);

	int N; cin >> N;
	ll A, B; cin >> A >> B;
	for (int i = 0; i < N; ++i) cin >> p[i];

	memset(DP, '?', sizeof DP);
	DP[0][0] = 0;
	for (int i = 0; i < N; ++i) {
		for (int j = 0; j <= N; ++j) {
			if (p[i] > j) DP[i+1][j] = min(DP[i][j]+A, DP[i+1][j]), DP[i+1][p[i]] = min(DP[i][j], DP[i+1][p[i]]);
			else DP[i+1][j] = min(DP[i][j]+B, DP[i+1][j]);
		}
	}

	ll ans = 1e18;
	for (int i = 1; i <= N; ++i) ans = min(DP[N][i], ans);
	cout << ans;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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.

Examples

Input

3 20 30
3 1 2


Output

20


Input

4 20 30
4 2 3 1


Output

50


Input

1 10 10
1


Output

0


Input

4 1000000000 1000000000
4 3 2 1


Output

3000000000


Input

9 40 50
5 3 4 7 6 1 2 9 8


Output

220

### Response

```cpp
#include <bits/stdc++.h>
#define f first
#define s second
using namespace std;
using ll = long long;
using ii = pair<int, int>;
constexpr int MN = 5005;

ll p[MN], DP[MN][MN];

int main() {
    if (fopen("in", "r")) freopen("in", "r", stdin), freopen("out", "w", stdout);
    ios_base::sync_with_stdio(0), cin.tie(0);

	int N; cin >> N;
	ll A, B; cin >> A >> B;
	for (int i = 0; i < N; ++i) cin >> p[i];

	memset(DP, '?', sizeof DP);
	DP[0][0] = 0;
	for (int i = 0; i < N; ++i) {
		for (int j = 0; j <= N; ++j) {
			if (p[i] > j) DP[i+1][j] = min(DP[i][j]+A, DP[i+1][j]), DP[i+1][p[i]] = min(DP[i][j], DP[i+1][p[i]]);
			else DP[i+1][j] = min(DP[i][j]+B, DP[i+1][j]);
		}
	}

	ll ans = 1e18;
	for (int i = 1; i <= N; ++i) ans = min(DP[N][i], ans);
	cout << ans;
}

```