#include <bits/stdc++.h>
using namespace std;
#define finish(x) return cout << x << endl, 0
#define ll long long

const int N = 5001;

int n, A, B, cnt[N];
ll dp[N][N];
vector <int> a;

ll solve(int idx, int j){
    if(idx == n) return 0;
    ll &ret = dp[idx][j];
    if(ret != -1) return ret;
    if(a[idx] == j) return solve(idx + 1, j + 1);
    if(a[idx] < j) return solve(idx + 1, j);
    if(cnt[j]) return solve(idx, j + 1);
    ret = solve(idx, j + 1) + B;
    cnt[a[idx]]++;
    ret = min(ret, solve(idx + 1, j) + A);
    cnt[a[idx]]--;
    return ret;
}
int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    memset(dp, -1, sizeof dp);
    cin >> n >> A >> B;
    a.resize(n);
    for(auto &i : a) cin >> i;
    cout << solve(0, 1) << endl;
}

### Prompt

Generate 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;
#define finish(x) return cout << x << endl, 0
#define ll long long

const int N = 5001;

int n, A, B, cnt[N];
ll dp[N][N];
vector <int> a;

ll solve(int idx, int j){
    if(idx == n) return 0;
    ll &ret = dp[idx][j];
    if(ret != -1) return ret;
    if(a[idx] == j) return solve(idx + 1, j + 1);
    if(a[idx] < j) return solve(idx + 1, j);
    if(cnt[j]) return solve(idx, j + 1);
    ret = solve(idx, j + 1) + B;
    cnt[a[idx]]++;
    ret = min(ret, solve(idx + 1, j) + A);
    cnt[a[idx]]--;
    return ret;
}
int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    memset(dp, -1, sizeof dp);
    cin >> n >> A >> B;
    a.resize(n);
    for(auto &i : a) cin >> i;
    cout << solve(0, 1) << endl;
}

```