#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int n,a,b;
ll mem[5005][5005];
int p[5005], pos[5005];
ll dp(int id, int cur){
    if (mem[id][cur] != -1) return mem[id][cur];
    if (id == 0) return mem[id][cur] = 0;
    if (pos[id] > cur) {
        //printf("%d at %d needs to move to %d, cost %d\n",id,pos[id],cur,b);
        return mem[id][cur] = dp(id-1,cur)+b; ///move id to cur.
    }
    //printf("%d at %d, can move to %d, cost %d\n",id,pos[id],cur,a);
    return mem[id][cur] =  min(dp(id-1,pos[id]),dp(id-1,cur)+a);;
}
int main(){
    scanf("%d%d%d",&n,&a,&b);
    for (int i = 1; i <= n; i++){
        scanf("%d",&p[i]);
        pos[p[i]] = i;
    }
    memset(mem,-1,sizeof(mem));
    printf("%lld",dp(n,n));

}

### Prompt

Please formulate 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;
typedef long long ll;
int n,a,b;
ll mem[5005][5005];
int p[5005], pos[5005];
ll dp(int id, int cur){
    if (mem[id][cur] != -1) return mem[id][cur];
    if (id == 0) return mem[id][cur] = 0;
    if (pos[id] > cur) {
        //printf("%d at %d needs to move to %d, cost %d\n",id,pos[id],cur,b);
        return mem[id][cur] = dp(id-1,cur)+b; ///move id to cur.
    }
    //printf("%d at %d, can move to %d, cost %d\n",id,pos[id],cur,a);
    return mem[id][cur] =  min(dp(id-1,pos[id]),dp(id-1,cur)+a);;
}
int main(){
    scanf("%d%d%d",&n,&a,&b);
    for (int i = 1; i <= n; i++){
        scanf("%d",&p[i]);
        pos[p[i]] = i;
    }
    memset(mem,-1,sizeof(mem));
    printf("%lld",dp(n,n));

}

```