#include <bits/stdc++.h>

typedef long long int int64;
int64 dp[5003][5003],dp2[5003];
int64 a,b;
int n;
int p[5003],ip[5003];
int main()
{
	scanf("%d%lld%lld",&n,&a,&b);
	for(int i = 1;i <= n;++i) {
		scanf("%d",&p[i]);
		ip[p[i]] = i;
	}
	memset(dp,-1,sizeof(dp));
	dp[0][0] = 0;
	dp2[0] = 0;
	for(int i = 1;i <= n;++i) {
		// (j,j+1)
		int64 min = dp[i-1][0];
		for(int j = 0;j <= n;++j) {
			if(dp[i-1][j]!=-1 && dp[i-1][j]<min) min=dp[i-1][j];
			if(ip[i-1]==j && dp2[i-1]<min) min=dp2[i-1];
			dp[i][j] = min + (ip[i] <= j ? a : b);
			if(ip[i] == j+1) dp2[i] = min;
		}
	}
	int64 ans = dp2[n];
	for(int i = 0;i <= n;++i) if(dp[n][i] < ans) ans = dp[n][i];
	printf("%lld\n", ans);
}

### 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>

typedef long long int int64;
int64 dp[5003][5003],dp2[5003];
int64 a,b;
int n;
int p[5003],ip[5003];
int main()
{
	scanf("%d%lld%lld",&n,&a,&b);
	for(int i = 1;i <= n;++i) {
		scanf("%d",&p[i]);
		ip[p[i]] = i;
	}
	memset(dp,-1,sizeof(dp));
	dp[0][0] = 0;
	dp2[0] = 0;
	for(int i = 1;i <= n;++i) {
		// (j,j+1)
		int64 min = dp[i-1][0];
		for(int j = 0;j <= n;++j) {
			if(dp[i-1][j]!=-1 && dp[i-1][j]<min) min=dp[i-1][j];
			if(ip[i-1]==j && dp2[i-1]<min) min=dp2[i-1];
			dp[i][j] = min + (ip[i] <= j ? a : b);
			if(ip[i] == j+1) dp2[i] = min;
		}
	}
	int64 ans = dp2[n];
	for(int i = 0;i <= n;++i) if(dp[n][i] < ans) ans = dp[n][i];
	printf("%lld\n", ans);
}

```