#include <bits/stdc++.h>

using namespace std;

const int Nmax = 5005;
typedef long long ll;

int n, A, B, i, j, p;
int smaller[Nmax], a[Nmax];

int main()
{
//    freopen("input", "r", stdin);
    cin.sync_with_stdio(false);

    cin >> n >> B >> A;
    for(i=1; i<=n; ++i) cin >> a[i];

    vector<ll> dp(n+2), new_dp(n+2);

    for(i=1; i<=n; ++i)
    {
        for(p=1; p<=n; ++p)
            if(a[p] == i) break;

        for(j=p; j<=n; ++j)
            smaller[j]++;

        for(j=1; j<p; ++j) new_dp[j] = dp[j];

        for(j=p; j<=n; ++j)
            new_dp[j] = min(dp[p-1] + (ll) A * (smaller[j] - smaller[p]), dp[j] + B);

        swap(dp, new_dp);
    }

    cout << dp[n] << '\n';
    return 0;
}

### 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;

const int Nmax = 5005;
typedef long long ll;

int n, A, B, i, j, p;
int smaller[Nmax], a[Nmax];

int main()
{
//    freopen("input", "r", stdin);
    cin.sync_with_stdio(false);

    cin >> n >> B >> A;
    for(i=1; i<=n; ++i) cin >> a[i];

    vector<ll> dp(n+2), new_dp(n+2);

    for(i=1; i<=n; ++i)
    {
        for(p=1; p<=n; ++p)
            if(a[p] == i) break;

        for(j=p; j<=n; ++j)
            smaller[j]++;

        for(j=1; j<p; ++j) new_dp[j] = dp[j];

        for(j=p; j<=n; ++j)
            new_dp[j] = min(dp[p-1] + (ll) A * (smaller[j] - smaller[p]), dp[j] + B);

        swap(dp, new_dp);
    }

    cout << dp[n] << '\n';
    return 0;
}

```