#include <cstdio>
#include <algorithm>

using namespace std;

int p[5000];
long long dp[5001][5001];

int main() {
    int n, a, b, i, j;
    long long ans = 1e18;
    
    scanf("%d %d %d", &n, &a, &b);
    
    for (i = 0; i < n; i++) scanf("%d", &p[i]);
    
    for (i = 0; i < n; i++) {
        int x = p[i];
        
        for (j = 0; j < x; j++) dp[i + 1][j] = dp[i][j] + a;
        for (j = x + 1; j <= n; j++) dp[i + 1][j] = dp[i][j] + b;
        
        dp[i + 1][x] = 1e18;
        for (j = 0; j < x; j++) dp[i + 1][x] = min(dp[i + 1][x], dp[i][j]);
    }
    
    for (i = 0; i <= n; i++) ans = min(ans, dp[n][i]);
    
    printf("%lld\n", ans);
    
    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 <cstdio>
#include <algorithm>

using namespace std;

int p[5000];
long long dp[5001][5001];

int main() {
    int n, a, b, i, j;
    long long ans = 1e18;
    
    scanf("%d %d %d", &n, &a, &b);
    
    for (i = 0; i < n; i++) scanf("%d", &p[i]);
    
    for (i = 0; i < n; i++) {
        int x = p[i];
        
        for (j = 0; j < x; j++) dp[i + 1][j] = dp[i][j] + a;
        for (j = x + 1; j <= n; j++) dp[i + 1][j] = dp[i][j] + b;
        
        dp[i + 1][x] = 1e18;
        for (j = 0; j < x; j++) dp[i + 1][x] = min(dp[i + 1][x], dp[i][j]);
    }
    
    for (i = 0; i <= n; i++) ans = min(ans, dp[n][i]);
    
    printf("%lld\n", ans);
    
    return 0;
}

```