#include <bits/stdc++.h>

using namespace std;
const int N = 5e3 + 5;
int a[N], p[N];
long long dp[N][N], mn[N][N], A, B;

int main() {
    int n; cin >> n >> A >> B;
    for (int i = 1; i <= n; i++) cin >> a[i], p[a[i]] = i;
    memset(dp, 63, sizeof dp);
    for (int x = 0; x <= n; x++) mn[0][x] = dp[0][x] = 0;
    for (int i = 1; i <= n; i++) {
        for (int x = 0; x <= n; x++) {
            dp[i][x] = (p[i] <= x ? A : B) + mn[i - 1][x];
            if (p[i] == x) dp[i][x] = min(dp[i][x], mn[i - 1][x - 1]);
            else if (x) dp[i][x] = min(dp[i][x], (p[i] < x ? A : B) + mn[i - 1][x - 1]);
            mn[i][x] = min((x ? mn[i][x - 1] : 1ll<<62), dp[i][x]);
        }
    }
    cout << mn[n][n] << "\n";

    return 0;
}

### 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>

using namespace std;
const int N = 5e3 + 5;
int a[N], p[N];
long long dp[N][N], mn[N][N], A, B;

int main() {
    int n; cin >> n >> A >> B;
    for (int i = 1; i <= n; i++) cin >> a[i], p[a[i]] = i;
    memset(dp, 63, sizeof dp);
    for (int x = 0; x <= n; x++) mn[0][x] = dp[0][x] = 0;
    for (int i = 1; i <= n; i++) {
        for (int x = 0; x <= n; x++) {
            dp[i][x] = (p[i] <= x ? A : B) + mn[i - 1][x];
            if (p[i] == x) dp[i][x] = min(dp[i][x], mn[i - 1][x - 1]);
            else if (x) dp[i][x] = min(dp[i][x], (p[i] < x ? A : B) + mn[i - 1][x - 1]);
            mn[i][x] = min((x ? mn[i][x - 1] : 1ll<<62), dp[i][x]);
        }
    }
    cout << mn[n][n] << "\n";

    return 0;
}

```