#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long LL;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL N = 1000000007;

LL n,l,r;
LL a[5005],dp[5005];

int main(){
	memset(dp,0x3f,sizeof(dp));
	cin >> n >> r >> l;
	for(LL i = 1;i <= n;i ++){
		cin >> a[i];
		a[i] ++;
	}
	a[0] = 1; dp[0] = 0; a[n + 1] = n + 2;
	for(LL i = 1;i <= n + 1;i ++){
		LL sr = 0,sl = 0;
		for(LL j = i - 1;j >= 0;j --){
			if(a[j] < a[i]) dp[i] = min(dp[i],dp[j] + r * sr + l * sl);
			if(a[j] > a[i]) sr ++; if(a[j] < a[i]) sl ++;
		}
	}
	cout << dp[n + 1] << endl;
	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 <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long LL;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL N = 1000000007;

LL n,l,r;
LL a[5005],dp[5005];

int main(){
	memset(dp,0x3f,sizeof(dp));
	cin >> n >> r >> l;
	for(LL i = 1;i <= n;i ++){
		cin >> a[i];
		a[i] ++;
	}
	a[0] = 1; dp[0] = 0; a[n + 1] = n + 2;
	for(LL i = 1;i <= n + 1;i ++){
		LL sr = 0,sl = 0;
		for(LL j = i - 1;j >= 0;j --){
			if(a[j] < a[i]) dp[i] = min(dp[i],dp[j] + r * sr + l * sl);
			if(a[j] > a[i]) sr ++; if(a[j] < a[i]) sl ++;
		}
	}
	cout << dp[n + 1] << endl;
	return 0;
}
```