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
# -*- coding: utf-8 -*-
"""
Created on Sat Jun 15 22:01:38 2019

@author: aozora
"""
N, *L = map(int, open(0).read().split())
MOD = 10**9+7
graph = [[] for i in range(N+1)]
for a, b in zip(*[iter(L)]*2):
  graph[a].append(b)
  graph[b].append(a)
root = 1
parent = [0]*(N+1)
order = []
stack = [root]
while stack:
  x = stack.pop()
  order.append(x)
  for c in graph[x]:
    if parent[x]==c:
      continue
    stack.append(c)
    parent[c] = x

fact = [1] * (N + 10)
for n in range(1, N + 10):
    fact[n] = n * fact[n - 1] % MOD
fact_inv = [1] * (N + 10)
fact_inv[-1] = pow(fact[-1], MOD - 2, MOD)
for n in range(N + 9, 0, -1):
    fact_inv[n - 1] = fact_inv[n] * n % MOD

dp_d = [1]*(N+1)
size_d = [0]*(N+1)
for v in order[::-1]:
  s = size_d[v]
  p = parent[v]
  dp_d[v] *= fact[s]
  dp_d[v] %= MOD
  m = dp_d[v]
  size_d[p] += s+1
  dp_d[p] *= fact_inv[s+1]*m
  dp_d[p] %= MOD

size_u = [N-2-x for x in size_d]
dp_u = [1]*(N+1)

for v in order[1:]:
  p = parent[v]
  x = dp_d[p]
  x *= fact_inv[size_d[p]]
  x *= fact[size_d[v]+1]
  x *= pow(dp_d[v],MOD-2,MOD)
  x *= fact[size_u[v]]
  x *= fact_inv[size_u[p]+1]
  x *= dp_u[p]
  dp_u[v] = x%MOD

for xd, xu, sd, su in zip(dp_d[1:], dp_u[1:], size_d[1:], size_u[1:]):
    su += 1
    x = xd * xu * fact[sd + su] * fact_inv[sd] * fact_inv[su] % MOD
    print(x)