#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll INFL = 1e18;
const int MN = 5050;
ll dp[MN][MN];
ll dt[MN];
ll ord[MN];
int main() {
	ios::sync_with_stdio(0);cin.tie(0);
	ll n,a,b;
	cin >> n >> a >> b;
	for(int i=0;i<n;i++) {
		int t;
		cin >> t;
		t--;
		ord[t] = i+1;
	}
	memset(dp,0,sizeof(dp));
	for(int i=0;i<n;i++) {
		ll ma = INFL;
		for(int j=0;j<=n;j++) {
			if(i > 0) {
				if(j == ord[i-1]) {
					ma = min(ma,dt[i-1]);
				}
				ma = min(ma,dp[i-1][j]);
			} else {
				ma = 0;
			}
			if(j == ord[i]) {
				dt[i] = ma;
			}
			dp[i][j] = ma + (j>=ord[i]?a:b);
		}
	}
	ll res = INFL;
	for(int i=0;i<=n;i++) {
		res = min(res,dp[n-1][i]);
	}
	res = min(res,dt[n-1]);
	cout << res << '\n';
}

### 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;
typedef long long ll;
const ll INFL = 1e18;
const int MN = 5050;
ll dp[MN][MN];
ll dt[MN];
ll ord[MN];
int main() {
	ios::sync_with_stdio(0);cin.tie(0);
	ll n,a,b;
	cin >> n >> a >> b;
	for(int i=0;i<n;i++) {
		int t;
		cin >> t;
		t--;
		ord[t] = i+1;
	}
	memset(dp,0,sizeof(dp));
	for(int i=0;i<n;i++) {
		ll ma = INFL;
		for(int j=0;j<=n;j++) {
			if(i > 0) {
				if(j == ord[i-1]) {
					ma = min(ma,dt[i-1]);
				}
				ma = min(ma,dp[i-1][j]);
			} else {
				ma = 0;
			}
			if(j == ord[i]) {
				dt[i] = ma;
			}
			dp[i][j] = ma + (j>=ord[i]?a:b);
		}
	}
	ll res = INFL;
	for(int i=0;i<=n;i++) {
		res = min(res,dp[n-1][i]);
	}
	res = min(res,dt[n-1]);
	cout << res << '\n';
}

```