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 fractions import gcd
import sys
sys.setrecursionlimit(4100000)

def os(): return input()
def oi(): return int(input())
def oli(): return map(int, input().split())
def olai(): return list(map(int, input().split()))
def olas(): return list(input().split())
def mlai(L): return [int(input()) for _ in range(L)]
def mlas(L): return [input() for _ in range(L)]
def ar(a, b, c=-1, d=-1, INIT=0):
    if(c==-1): return [[INIT]*b for _ in range(a)]
    if(d==-1): return [[[INIT]*c for _ in range(b)] for __ in range(a)]
    return [[[[INIT]*d for _ in range(c)] for __ in range(b)] for ___ in range(a)]

def lcm(a, b): return (a*b)//gcd(a, b)
#MOD
P = 10**9 + 7
N = 200001 #使う最大値+1以上にする、値に注意3*10^5とかにしとくと安心
inv = [0] + [1] # 1/x
finv = [1] + [1] # 1/x!
fac = [1] + [1] # x!
for i in range(2, N):
  inv += [inv[P % i] * (P - int(P / i)) % P]
  fac += [(fac[i-1] * i) % P]
  finv += [(finv[i-1] * inv[i]) % P]

def comb(a, b):
    if a<b or b < 0: return 0 # error
    return (fac[a] * ((finv[b] * finv[a-b]) % P)) %P

#全方位木dp
#-------------------------------------------------------------------
#扱うクラス
class T:
    num = 1
    child = 0

    def __init__(self, a, b):
        self.num = a
        self.child = b

#零元
zero = T(1, 0)
#一個上に遷移した時の変化
def step(x):
    return T(x.num, x.child+1)
#マージ
def merge(x, y):
    return T((x.num*y.num*comb(x.child+y.child, x.child))%P, x.child+y.child)

G = [[] for _ in range(200001)]
down_data = [zero for _ in range(200001)]
dat = [zero for _ in range(200001)]

def dfs(now, parent):
    cnt = 0
    for to in G[now]:
        if(to==parent): continue
        cnt += 1
        dfs(to, now)
        down_data[now] = merge(down_data[now], step(down_data[to]))

def dfs_rev(now, parent, rev):
    #累積
    left = []
    right = []
    cnt = 0
    for to in G[now]:
        if(to==parent): continue
        if(len(left)==0): left.append(step(down_data[to]))
        else: left.append(merge(left[cnt-1], step(down_data[to])))
        cnt += 1
    cnt = 0
    for to in reversed(G[now]):
        if(to==parent): continue
        if(len(right)==0): right.append(step(down_data[to]))
        else: right.append(merge(right[cnt-1], step(down_data[to])))
        cnt += 1
    cnt = 0
    dat[now] = merge(rev, down_data[now])

    for to in G[now]:
        if(to==parent): continue
        x = zero if cnt==0 else left[cnt-1]
        y = zero if cnt==len(right)-1 else right[len(right)-2-cnt]
        dfs_rev(to, now, step(merge(rev, merge(x, y))))
        cnt += 1
#-------------------------------------------------------------------
def main():
    n = oi()
    for i in range(n-1):
        a, b = oli()
        G[a].append(b)
        G[b].append(a)
    root = -1
    for i in range(1, n+1):
        if(len(G[i]) == 1):
            root = i
            break
    dfs(root, -1)
    dfs_rev(root, -1, zero)
    for i in range(1, n+1):
        print(dat[i].num)

if __name__ == "__main__": main()
