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
sys.setrecursionlimit(1234567890)
from collections import deque

n, *L = map(int, open(0).read().split())
mod = 10 ** 9 + 7


def prepare(m, mod=10 ** 9 + 7):
	fac = [1] * (m + 1)
	inv = [1] * (m + 1)
	for i in range(1, m + 1):
		fac[i] = fac[i - 1] * i % mod
	inv[-1] = pow(fac[-1], mod - 2, mod)
	for i in range(m - 1, -1, -1):
		inv[i] = inv[i + 1] * (i + 1) % mod
	return fac, inv


fac, inv = prepare(n)

con = [[] for _ in range(n)]
for s, t in zip(*[iter(L)] * 2):
	con[s - 1] += t - 1,
	con[t - 1] += s - 1,

par = [-1] * n
q = deque([0])
odr = []
while q:
	cur = q.popleft()
	odr += cur,
	for nxt in con[cur]:
		if nxt != par[cur]:
			par[nxt] = cur
			q.append(nxt)

sz = [1] * (n + 1)
st = 1
for i in odr[::-1]:
	sz[par[i]] += sz[i]
	st *= pow(sz[i], mod - 2, mod)
	st %= mod

# inv[n-1]*fac[sz[i]]*inv[sz[i]-1]*inv[n-sz[i]]*fac[n-sz[i]-1]*fac[n-1]=sz[i]*(n-sz[i])^-1
ans = [st * fac[n] % mod] + [0] * (n - 1)
for i in odr[1:]:
	ans[i] = ans[par[i]] * sz[i] % mod * pow(n - sz[i], mod - 2, mod) % mod
print(*ans)
