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
n = int(input())
big = 1
mod = 10 ** 9 + 7
for i in range(1,n):
  big *= i
  big %= mod
def inv(x):
  y = 1
  while x != 1:
    y *= mod // x + 1
    y %= mod
    x -= mod % x
  return y
tribe = [1 for _ in range(n)]
we = [1 for _ in range(n)]
peer = [[] for _ in range(n)]
for _ in range(n-1):
  a,b = map(int,input().split())
  a -= 1
  b -= 1
  peer[a].append(b)
  peer[b].append(a)
seen = [0 for _ in range(n)]
pre = [0 for _ in range(n)]
seen[0] = 1
rank = []
now = [0]
while now:
  rank.append(now)
  last = now
  now = []
  for x in last:
    for y in peer[x]:
      if seen[y] == 0:
        seen[y] = 1
        now.append(y)
        pre[y] = x
#print(seen)
#print(pre)
#print(rank)
pre[0] = 10**6
rank.reverse()
for z in rank:
  if z == [0]:
    continue
  for w in z:
    tribe[pre[w]] += tribe[w]
    we[pre[w]] *= tribe[w] * we[w]
    we[pre[w]] %= mod
#print(tribe)
#print(we)
rank.reverse()
for z in rank:
  if z == [0]:
    continue
  for w in z:
    tmp = we[pre[w]]
    tmp *= inv(we[w]*tribe[w]%mod)
    tmp *= n - tribe[w]
    we[w] *= tmp
    we[w] %= mod
#print(we)
for i in range(n):
  print(big*inv(we[i])%mod)