Problem:
 Score : 600 points    Problem Statement  We have a tree with N vertices numbered 1 to N . The i -th edge in this tree connects Vertex a_i and b_i .
For each k=1, ..., N , solve the problem below:   Consider writing a number on each vertex in the tree in the following manner:  First, write 1 on Vertex k .  Then, for each of the numbers 2, ..., N in this order, write the number on the vertex chosen as follows:  Choose a vertex that still does not have a number written on it and is adjacent to a vertex with a number already written on it. If there are multiple such vertices, choose one of them at random.      Find the number of ways in which we can write the numbers on the vertices, modulo (10^9+7) .       Constraints   2 \leq N \leq 2 \times 10^5  1 \leq a_i,b_i \leq N  The given graph is a tree.         Input  Input is given from Standard Input in the following format:  N  a_1  b_1  :  a_{N-1}  b_{N-1}       Output  For each k=1, 2, ..., N in this order, print a line containing the answer to the problem.        Sample Input 1 3
1 2
1 3      Sample Output 1 2
1
1  The graph in this input is as follows:   For k=1 , there are two ways in which we can write the numbers on the vertices, as follows:   Writing 1, 2, 3 on Vertex 1, 2, 3 , respectively  Writing 1, 3, 2 on Vertex 1, 2, 3 , respectively        Sample Input 2 2
1 2      Sample Output 2 1
1  The graph in this input is as follows:        Sample Input 3 5
1 2
2 3
3 4
3 5      Sample Output 3 2
8
12
3
3  The graph in this input is as follows:        Sample Input 4 8
1 2
2 3
3 4
3 5
3 6
6 7
6 8      Sample Output 4 40
280
840
120
120
504
72
72  The graph in this input is as follows:

Solution (Python):
import sys

stdin = sys.stdin

ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
ns = lambda: stdin.readline().rstrip()  

n = ni()
g = []
for i in range(n): g.append([])
for i in range(n-1):
    x, y = na()
    x -= 1
    y -= 1
    g[x].append(y)
    g[y].append(x)

dp = [0] * n
ep = [0] * n
des = [0] * n

def enumfif(n, mod):
    f = [0] * (n+1)
    invf = [0] * (n+1)
    f[0] = 1
    for i in range(1, n+1):
        f[i] = f[i-1] * i % mod

    a = f[n]
    b = mod
    p, q = 1, 0
    while b > 0:
        c = a//b
        d = a; a = b; b = d % b
        d = p; p = q; q = d-c*q
    invf[n] = p + mod if p < 0 else p
    for i in range(n-1,-1,-1):
        invf[i] = invf[i+1] * (i+1) % mod
    return f, invf

def C(n, r, mod, fif):
    if n < 0 or r < 0 or r > n:
        return 0
    return fif[0][n] * fif[1][r] * fif[1][n-r] % mod

mod = 1000000007
fif = enumfif(200005, mod)

sys.setrecursionlimit(200005)

def dfs(cur, pre, g):
    val = 1
    de = 0
    for e in g[cur]:
        if e == pre: continue
        dfs(e, cur, g)
        de += des[e]
        val = val * dp[e] * C(de, des[e], mod, fif) % mod
    dp[cur] = val
    des[cur] = de + 1

def dfs2(cur, pre, g):
    
    for e in g[cur]:
        if e == pre: continue
        val = dp[cur] * fif[1][n-1] * fif[0][n-des[e]-1] * fif[0][des[e]] % mod
        dp[e] = val * C(n-1, n-des[e], mod, fif) % mod
        
        
        dfs2(e, cur, g)

def invl(a, mod):
    b = mod
    p = 1; q = 0
    while b > 0:
        c = a // b
        a, b = b, a%b
        p, q = q, p-c*q
    return p + mod if p < 0 else p


dfs(0, -1, g)
dfs2(0, -1, g)
for val in dp:
    print(val)