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
#!/usr/bin/env python3

import sys
import math
from bisect import bisect_right as br
from bisect import bisect_left as bl
sys.setrecursionlimit(2147483647)
from heapq import heappush, heappop,heappushpop
from collections import defaultdict
from itertools import accumulate
from collections import Counter
from collections import deque
from operator import itemgetter
from itertools import permutations
mod = 10**9+7
inf = float('inf')
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))

def dfs_1(s,p): #部分木のサイズを数えるdfs
    par[s] = p
    for v in edges[s]:
        if v == p:
            continue
        size[s] += dfs_1(v,s)
    return size[s]

def dfs_2(s,p): #1を根としたときに進む向きのdp計算用dfs
    dp1[s] = fact[size[s]-1]
    for v in edges[s]:
        if v == p:
            continue
        dp1[s] = (dp1[s]*dfs_2(v,s))%mod
        dp1[s] = (dp1[s]*pow(fact[size[v]], mod-2, mod))%mod
    return dp1[s]

def comb(n,r):
    return fact[n]*pow(fact[r],mod-2,mod)*pow(fact[n-r],mod-2,mod) % mod

def bfs(s):
    que = deque()
    que.append((s,-1))
    while que:
        v,p = que.popleft()
        if v == 0:
            dp2[v] = dp1[v]
        else:
            inv = (dp2[p]*pow((dp1[v]*comb(n-1,size[v]))%mod,mod-2,mod)) % mod
            dp2[v] = (dp1[v]*inv*comb(n-1,size[v]-1)) % mod
        for u in edges[v]:
            if u == p:
                continue
            que.append((u,v))
n = I()
fact = [1]*(n+1)
par = [None]*n
for i in range(1,n+1):
    fact[i] = i*fact[i-1]
    fact[i] %= mod

edges = [[] for _ in range(n)]
size = [1]*n
dp1 = [0]*n #1を根としたときに進む向きのdp
for _ in range(n-1):
    a,b = LI()
    edges[a-1].append(b-1)
    edges[b-1].append(a-1)

dfs_1(0,-1)
dfs_2(0,-1)
dp2 = [0]*n
bfs(0)
for i in dp2:
    print(i)


