Problem:
Score : 1000 points

Problem Statement

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.

Sample Input 1

3 20 30
3 1 2

Sample Output 1

20

Shifting (p_1, p_2, p_3) to the left by one results in p = (1, 2, 3).

Sample Input 2

4 20 30
4 2 3 1

Sample Output 2

50

One possible sequence of operations is as follows:

Shift (p_1, p_2, p_3, p_4) to the left by one. Now we have p = (2, 3, 1, 4).

Shift (p_1, p_2, p_3) to the right by one. Now we have p = (1, 2, 3, 4).

Here, the total cost is 20 + 30 = 50.

Sample Input 3

1 10 10
1

Sample Output 3

0

Sample Input 4

4 1000000000 1000000000
4 3 2 1

Sample Output 4

3000000000

Sample Input 5

9 40 50
5 3 4 7 6 1 2 9 8

Sample Output 5

220

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

// KALAM
# include<bits/stdc++.h>

using namespace std;

const int N = 5000 + 77;
int n , X , Y;
int a[N] , b[N] , F[N];
long long dp[N][N];
int main() {
   memset(dp , 63 , sizeof(dp));
   for(int i = 0;i < N;++ i)
      dp[0][i] = 0;
   scanf("%d" , & n);
   scanf("%d %d" , & X , & Y);
   for(int i = 1;i <= n;++ i)
      scanf("%d" , a + i);
   for(int i = 1;i <= n;++ i) {
      for(int j = i;j <= n;++ j)
         dp[i][j] = 0;
      for(int j = 1;j < i;++ j)
         F[i] += (a[j] > a[i]);
      b[i] = i;
      for(int j = i;j > 1;-- j)
         if(a[b[j]] < a[b[j - 1]])
            swap(b[j] , b[j - 1]);
      for(int j = 2;j <= i;++ j) {
         int t = i - j + 1;
         int pos = b[i - t + 1];
         int leftmore = F[pos];
         int rightmore = t - 1 - F[pos];
         int rightless = i - pos - rightmore;
         dp[i][t] = dp[pos - 1][leftmore + 1] + rightless * 1ll * Y;
         if(rightless == 0)
            dp[i][t] = min(dp[i][t] , dp[i][t + 1]);
         else
            dp[i][t] = min(dp[i][t] , dp[i][t + 1] + X);
      }
   }

   printf("%lld\n" , dp[n][1]);
   return 0;
}