PermuTree (easy version)
This is the easy version of the problem. The differences between the two versions are the constraint on $$$n$$$ and the time limit. You can make hacks only if both versions of the problem are solved.You are given a tree with $$$n$$$ vertices rooted at vertex $$$1$$$.For some permutation$$$^\dagger$$$ $$$a$$$ of length $$$n$$$, let $$$f(a)$$$ be the number of pairs of vertices $$$(u, v)$$$ such that $$$a_u < a_{\operatorname{lca}(u, v)} < a_v$$$. Here, $$$\operatorname{lca}(u,v)$$$ denotes the lowest common ancestor of vertices $$$u$$$ and $$$v$$$.Find the maximum possible value of $$$f(a)$$$ over all permutations $$$a$$$ of length $$$n$$$.$$$^\dagger$$$ A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array), and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).
The tree in the first test:   One possible optimal permutation $$$a$$$ is $$$[2, 1, 4, 5, 3]$$$ with $$$4$$$ suitable pairs of vertices:   $$$(2, 3)$$$, since $$$\operatorname{lca}(2, 3) = 1$$$ and $$$1 < 2 < 4$$$,  $$$(2, 4)$$$, since $$$\operatorname{lca}(2, 4) = 1$$$ and $$$1 < 2 < 5$$$,  $$$(2, 5)$$$, since $$$\operatorname{lca}(2, 5) = 1$$$ and $$$1 < 2 < 3$$$,  $$$(5, 4)$$$, since $$$\operatorname{lca}(5, 4) = 3$$$ and $$$3 < 4 < 5$$$. The tree in the third test:   The tree in the fourth test:   
The first line contains a single integer $$$n$$$ ($$$2 \le n \le 5000$$$).The second line contains $$$n - 1$$$ integers $$$p_2,p_3,\ldots,p_n$$$ ($$$1 \le p_i < i$$$) indicating that there is an edge between vertices $$$i$$$ and $$$p_i$$$.
Output the maximum value of $$$f(a)$$$.
from sys import stdin
input = lambda : stdin.readline()[:-1]

def calc(a):
    sm = sum(a)
    dp = 1
    for i in a:
        dp |= dp << i
    res = 0
    for j in range(sm // 2 + 2):
        if dp >> j & 1:
            res = max(res, j * (sm - j))
    return res

def solve():
    n = int(input())
    p = [-1] + list(map(lambda x: int(x) - 1, input().split()))
    dp = [1] * n
    chi = [[] for i in range(n)]
    for i in range(n - 1, 0, -1):
        chi[p[i]].append(dp[i])
        dp[p[i]] += dp[i]
    ans = 0
    for v in range(n):
        if len(chi[v]) <= 1:
            continue
        ans += calc(chi[v])
    print(ans)
solve()
from sys import stdin
input = lambda : stdin.readline()[:-1]

def calc(a):
    sm = sum(a)
    dp = 1
    for i in a:
        dp |= dp << i
    res = 0
    for j in range(sm // 2 + 2):
        if dp >> j & 1:
            res = max(res, j * (sm - j))
    return res

def solve():
    n = int(input())
    p = [-1] + list(map(lambda x: int(x) - 1, input().split()))
    dp = [1] * n
    chi = [[] for i in range(n)]
    for i in range(n - 1, 0, -1):
        chi[p[i]].append(dp[i])
        dp[p[i]] += dp[i]
    ans = 0
    for v in range(n):
        if len(chi[v]) <= 1:
            continue
        ans += calc(chi[v])
    print(ans)
solve()
from sys import stdin
input = lambda : stdin.readline()[:-1]

def calc(a):
    sm = sum(a)
    dp = 1
    for i in a:
        dp |= dp << i
    res = 0
    for j in range(sm // 2 + 2):
        if dp >> j & 1:
            res = max(res, j * (sm - j))
    return res

def solve():
    n = int(input())
    p = [-1] + list(map(lambda x: int(x) - 1, input().split()))
    dp = [1] * n
    chi = [[] for i in range(n)]
    for i in range(n - 1, 0, -1):
        chi[p[i]].append(dp[i])
        dp[p[i]] += dp[i]
    ans = 0
    for v in range(n):
        if len(chi[v]) <= 1:
            continue
        ans += calc(chi[v])
    print(ans)
solve()