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
sys.setrecursionlimit(10**7)

U = 2*10**5
MOD = 10**9+7
 
fact = [1]*(U+1)
fact_inv = [1]*(U+1)
 
for i in range(1,U+1):
    fact[i] = (fact[i-1]*i)%MOD
fact_inv[U] = pow(fact[U], MOD-2, MOD)
 
for i in range(U,0,-1):
    fact_inv[i-1] = (fact_inv[i]*i)%MOD

n = int(input())
T = [[] for _ in range(n)]
for _ in range(n-1):
  a, b = map(int, input().split())
  a -= 1
  b -= 1
  T[a].append(b)
  T[b].append(a)
size = [0]*n
par = [-1]*n
dp1 = [-1]*n
dp2 = [-1]*n
def dfs1(p, v):
  if dp1[v] != -1:
    return dp1[v]
  temp = 1
  s = 1
  for nv in T[v]:
    if nv == p:
      continue
    temp *= dfs1(v, nv)
    temp *= fact_inv[size[nv]]
    temp %= MOD
    par[nv] = v
    s += size[nv]
  size[v] = s
  temp *= fact[s-1]
  temp %= MOD
  dp1[v] = temp
  return temp
dfs1(-1, 0)
def dfs2(v):
  if v == 0:
    return dp1[0]
  if dp2[v] != -1:
    return dp2[v]
  temp = dfs2(par[v])
  temp *= fact[n-size[v]-1]
  temp *= fact[size[v]]
  temp %= MOD
  temp *= fact_inv[size[v]-1]
  temp *= fact_inv[n-size[v]]
  temp %= MOD
  dp2[v] = temp
  return temp
for i in range(n):
  print(dfs2(i))