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
from collections import deque
N = int(input())
ab = [list(map(int,input().split())) for _ in range(N-1)]
mod = 10**9+7
fact = [1]*(N+1)
fact_inv = [1]*(N+1)
for i in range(1,N+1):
  fact[i] = (fact[i-1]*i)%mod
  
fact_inv[-1] = pow(fact[-1], mod-2, mod)
for i in range(N,0,-1):
  fact_inv[i-1] = fact_inv[i] * i % mod

g = [deque([]) for _ in range(N+1)]
for a,b in ab:
  g[a].append(b)
  g[b].append(a)
  
parents = [0]*(N+1)
q = deque([])
s = deque([1])
while s:
  x = s.pop()
  q.append(x)
  for y in g[x]:
    if y==parents[x]:
      continue
    else:
      parents[y] = x
      s.append(y)
      
q = list(q)
size_forward = [0]*(N+1)
dp_forward = [1]*(N+1)
for i in q[::-1]:
  p = parents[i]
  s = size_forward[i] + 1
  size_forward[p] += s
  dp_forward[i] *= fact[size_forward[i]]
  dp_forward[p] *= dp_forward[i] * fact_inv[s]
  dp_forward[p] %= mod
  
size_back = [N-1-i for i in size_forward]
dp_back = [1]*(N+1)
for i in q[1:]:
  p = parents[i]
  x = dp_back[p]
  x *= fact_inv[size_back[p]]
  x *= dp_forward[p]
  x *= fact_inv[size_forward[p]]
  x *= fact[size_forward[i]+1]
  x *= pow(dp_forward[i],mod-2,mod)
  x *= fact[size_back[i]-1]
  dp_back[i] = x % mod
  
for a,b,c,d in zip(size_forward[1:], dp_forward[1:], size_back[1:], dp_back[1:]):
  ans = b * d * fact[a+c] * fact_inv[a] * fact_inv[c]
  ans %= mod
  print(ans)