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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

N, A, B = map(int, input().split())
P = list(map(int, input().split()))
Pmin = [None]*N
Pmax = [None]*N
for i, p in enumerate(P):
    k = 20000
    for pj in P[:i]:
        if pj > p:
            k = min(k, pj)
    Pmin[p-1] = k

for i, p in enumerate(P):
    k = -1
    for pj in P[:i:-1]:
        if pj < p:
            k = max(k, pj)
    Pmax[p-1] = k
dp = [[10**15]*(N+2) for _ in range(N+2)]
dp[1][N] = 0
for k in range(N-1, -1, -1):
    for i in range(1, N - k + 1):
        x = dp[i][i+k]
        if Pmax[i+k-1] + 1 <= i:
            dp[i][i+k-1] = min(dp[i][i+k-1], x)
        else:
            dp[i][i+k-1] = min(dp[i][i+k-1], x+A)
        if Pmin[i-1] - 1 >= i+k:
            dp[i+1][i+k] = min(dp[i+1][i+k], x)
        else:
            dp[i+1][i+k] = min(dp[i+1][i+k], x+B)
print(min([dp[i][i] for i in range(1, N+1)]))