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(1000000)


def input():
    return sys.stdin.readline()


n = int(input())
e = [[] for _ in range(n)]
for _ in range(n-1):
    a, b = [int(i) - 1 for i in input().split()]
    e[a].append(b)
    e[b].append(a)

mod = 10**9+7

kkai = [1]
for i in range(1, 2*10 ** 5):
    kkai.append(kkai[-1] * i % mod)


def kai(x, p=mod):
    return kkai[x]


def comb(a, b, p=mod):
    if a < 0 or b < 0:
        return 0
    elif a < b:
        return 0
    if b == 0:
        return 1
    c = 1
    c *= kai(a, p)
    c *= pow(kai(b, p), p - 2, p)
    c *= pow(kai(a - b, p), p - 2, p)
    return c % p


child = [0] * n


def dfs1(i=0, r=-1):
    s = 1
    for j in e[i]:
        if j == r:
            continue
        s += dfs1(j, i)
    child[i] = s
    return s


dfs1()
a = [0] * n


def dfs2(i=0, r=-1):
    s = kai(child[i]-1)
    t = 1
    for j in e[i]:
        if j == r:
            continue
        s *= dfs2(j, i)
        s %= mod
        t *= kai(child[j])
        t %= mod
    s *= pow(t, mod - 2, mod)
    s %= mod
    a[i] = s
    return s


dfs2()
b = [0] * n


def dfs3(i=0, r=-1):
    if r == -1:
        b[i] = a[i]
    else:
        s = b[r] * child[i] % mod
        s *= pow(n-child[i], mod - 2, mod)
        s %= mod
        b[i] = s

    for j in e[i]:
        if j == r:
            continue
        dfs3(j, i)


dfs3()
print("\n".join(map(str, b)))
