#include <bits/stdc++.h>

using namespace std;


void solve() {
    int n, A, B;
    cin >> n >> A >> B;
    vector<int> a(n+2);
    a[0] = 0; a[n+1] = n+1;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    vector<long long> dp(n+2, 1ll<<60);
    dp[0] = 0;
    for (int i = 1; i <= n+1; i++) {
        int cnt = 0;
        for (int j = i-1; j >= 0; j--) {
            if (a[j] < a[i]) {
                dp[i] = min(dp[i], dp[j] + 1ll*cnt*A + 1ll*(i-1-j-cnt)*B);
            } else {
                cnt++;
            }
        }
    }
    cout << dp[n+1];
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    cout << endl;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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>

using namespace std;


void solve() {
    int n, A, B;
    cin >> n >> A >> B;
    vector<int> a(n+2);
    a[0] = 0; a[n+1] = n+1;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    vector<long long> dp(n+2, 1ll<<60);
    dp[0] = 0;
    for (int i = 1; i <= n+1; i++) {
        int cnt = 0;
        for (int j = i-1; j >= 0; j--) {
            if (a[j] < a[i]) {
                dp[i] = min(dp[i], dp[j] + 1ll*cnt*A + 1ll*(i-1-j-cnt)*B);
            } else {
                cnt++;
            }
        }
    }
    cout << dp[n+1];
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    cout << endl;
}

```