Problem:
Score : 800 points

Problem Statement

There is a tree with N vertices. The vertices are numbered 1 through N. The i-th edge (1 \leq i \leq N - 1) connects Vertex x_i and y_i. For vertices v and w (1 \leq v, w \leq N), we will define the distance between v and w d(v, w) as "the number of edges contained in the path v-w".

A squirrel lives in each vertex of the tree. They are planning to move, as follows. First, they will freely choose a permutation of (1, 2, ..., N), p = (p_1, p_2, ..., p_N). Then, for each 1 \leq i \leq N, the squirrel that lived in Vertex i will move to Vertex p_i.

Since they like long travels, they have decided to maximize the total distance they traveled during the process. That is, they will choose p so that d(1, p_1) + d(2, p_2) + ... + d(N, p_N) will be maximized. How many such ways are there to choose p, modulo 10^9 + 7?

Constraints

2 \leq N \leq 5,000

1 \leq x_i, y_i \leq N

The input graph is a tree.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
x_2 y_2
:
x_{N - 1} y_{N - 1}

Output

Print the number of the ways to choose p so that the condition is satisfied, modulo 10^9 + 7.

Sample Input 1

3
1 2
2 3

Sample Output 1

3

The maximum possible distance traveled by squirrels is 4.
There are three choices of p that achieve it, as follows:

(2, 3, 1)

(3, 1, 2)

(3, 2, 1)

Sample Input 2

4
1 2
1 3
1 4

Sample Output 2

11

The maximum possible distance traveled by squirrels is 6.
For example, p = (2, 1, 4, 3) achieves it.

Sample Input 3

6
1 2
1 3
1 4
2 5
2 6

Sample Output 3

36

Sample Input 4

7
1 2
6 3
4 5
1 7
1 5
2 3

Sample Output 4

396

Solution in Python (3.4.3) (judge verdict: Accepted)

import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 7)

import numpy as np
from scipy.sparse import csr_matrix
from scipy.sparse.csgraph import depth_first_order

MOD = 10**9 + 7

N = int(readline())
XY = list(map(int,read().split()))

graph = csr_matrix(([1]*(N-1),(XY[::2],XY[1::2])),(N+1,N+1))
dfs_order, par = depth_first_order(graph,1,directed=False)
dfs_order = dfs_order.tolist()
par = par.tolist()

subtree_size = [1] * (N+1)
is_cent = [True] * (N+1)

for v in dfs_order[::-1]:
    if subtree_size[v] < (N+1)//2:
        is_cent[v] = False
    p = par[v]
    if p < 0:
        break
    if subtree_size[v] > N//2:
        is_cent[p] = False
    subtree_size[p] += subtree_size[v]

centroid = [v for v in range(1,N+1) if is_cent[v]]
Ncent = len(centroid)

if Ncent == 1:
    c = centroid[0]
    sizes = [N-subtree_size[c]] + [subtree_size[v] for v in range(1,N+1) if par[v] == c]
else:
    sizes = [N//2,N//2]

def cumprod(arr,MOD):
    L = len(arr); Lsq = int(L**.5+1)
    arr = np.resize(arr,Lsq**2).reshape(Lsq,Lsq)
    for n in range(1,Lsq):
        arr[:,n] *= arr[:,n-1]; arr[:,n] %= MOD
    for n in range(1,Lsq):
        arr[n] *= arr[n-1,-1]; arr[n] %= MOD
    return arr.ravel()[:L]

def make_fact(U,MOD):
    x = np.arange(U,dtype=np.int64); x[0] = 1
    fact = cumprod(x,MOD)
    x = np.arange(U,0,-1,dtype=np.int64); x[0] = pow(int(fact[-1]),MOD-2,MOD)
    fact_inv = cumprod(x,MOD)[::-1]
    return fact,fact_inv

fact,fact_inv = make_fact(N+100,MOD)

def create_component_arr(n):
    x = np.full(n+1,fact[n],dtype=np.int64)
    x *= fact_inv[n::-1]; x %= MOD
    x *= x; x %= MOD
    x *= fact_inv[:n+1]; x %= MOD
    x[1::2] *= (-1); x %= MOD
    return x

def convolve(x,y):
    Lx = len(x); Ly = len(y)
    if Lx < Ly:
        Lx,Ly = Ly,Lx
        x,y = y,x
    z = np.zeros(Lx+Ly-1,np.int64)
    for n in range(Ly):
        z[n:n+Lx] += x*y[n]%MOD
    return z%MOD

x = np.array([1],np.int64)
for s in sizes:
    x = convolve(x,create_component_arr(s))

if Ncent == 1:
    answer = (x * fact[N:0:-1] % MOD).sum() % MOD
else:
    answer = (x * fact[N::-1] % MOD).sum() % MOD
print(answer)