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()))
q = [0] * n
for i, x in enumerate(p):
    q[x - 1] = i
dp = [[0] * (n + 1) for _ in range(n + 1)]
mi = [[n] * (n + 1) for _ in range(n + 1)]
ma = [[0] * (n + 1) for _ in range(n + 1)]
for i in range(n + 1):
    for j in range(i + 1, n + 1):
        mi[i][j] = min(mi[i][j - 1], q[j - 1])
        ma[i][j] = max(ma[i][j - 1], q[j - 1])
for i in range(1, n + 1):
    for j in range(i + 1):
        k = i - j
        s, t = 1, 1
        if j > 0 and j == p[mi[j - 1][n - k]]:
            s = 0
        if k > 0 and n - k + 1 == p[ma[j][n - k + 1]]:
            t = 0
        if j == 0:
            dp[j][k] = dp[j][k - 1] + a * t
        elif k == 0:
            dp[j][k] = dp[j - 1][k] + b * s
        else:
            dp[j][k] = min(dp[j - 1][k] + b * s, dp[j][k - 1] + a * t)
ans = float('inf')
for i in range(n + 1):
    ans = min(ans, dp[i][n - i])
print(ans)