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.

Examples

Input

3
1 2
1 3


Output

2
1
1


Input

2
1 2


Output

1
1


Input

5
1 2
2 3
3 4
3 5


Output

2
8
12
3
3


Input

8
1 2
2 3
3 4
3 5
3 6
6 7
6 8


Output

40
280
840
120
120
504
72
72
import sys
input = sys.stdin.readline

N = int(input())
vec = [[] for _ in range(N)]
for _ in range(N - 1) :
  a, b = map(int, input().split())
  vec[a - 1].append(b - 1)
  vec[b - 1].append(a - 1)

mod = 10 ** 9 + 7

fac = [1]
for i in range(1, N + 1) :
  fac.append(fac[-1] * i % mod)

inv = [pow(fac[-1], mod - 2, mod)]
for i in range(N, 0, -1) :
  inv.append(inv[-1] * i % mod)
inv = inv[::-1]

def comb(n, r) :
  return fac[n] * inv[n-r] * inv[r] % mod

pattern = [1] * N
subtree = [0] * N

# DFS
stack = [(0, -1, 0)]
while stack :
  cur, pre, op = stack.pop()
  
  if op == 0 :
    stack.append((cur, pre, 1))
    for nex in vec[cur] :
      if nex == pre : continue
      stack.append((nex, cur, 0))
  else :
    for nex in vec[cur] :
      pattern[cur] *= pattern[nex] * comb(subtree[cur] + subtree[nex], subtree[nex]) % mod
      pattern[cur] %= mod
      subtree[cur] += subtree[nex]
    subtree[cur] += 1

# BFS
queue = [(0, -1)]
while queue :
  cur, pre = queue.pop()
  for nex in vec[cur] :
    if nex == pre : continue
    pattern[nex] = pattern[cur] * subtree[nex] * inv[N - subtree[nex]] * fac[N - subtree[nex] - 1] % mod
    queue.append((nex, cur))
  
for i in range(N) :
  print(pattern[i])