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/ python3.8
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

N = int(readline())
m = map(int, read().split())
AB = zip(m, m)

MOD = 10 ** 9 + 7

graph = [[] for _ in range(N + 1)]
for a, b in AB:
    graph[a].append(b)
    graph[b].append(a)

root = 1
parent = [0] * (N + 1)
order = []
stack = [root]
while stack:
    x = stack.pop()
    order.append(x)
    for y in graph[x]:
        if y == parent[x]:
            continue
        parent[y] = x
        stack.append(y)

fact = [1] * (N + 10)
for n in range(1, N + 10):
    fact[n] = n * fact[n - 1] % MOD
fact_inv = [1] * (N + 10)
fact_inv[-1] = pow(fact[-1], MOD - 2, MOD)
for n in range(N + 9, 0, -1):
    fact_inv[n - 1] = fact_inv[n] * n % MOD

size_d = [0] * (N + 1)
dp_d = [1] * (N + 1)

for v in order[::-1]:
    dp_d[v] *= fact[size_d[v]]
    dp_d[v] %= MOD
    p = parent[v]
    s = size_d[v] + 1
    size_d[p] += s
    dp_d[p] *= fact_inv[s] * dp_d[v]
    dp_d[p] %= MOD

size_u = [N - 2 - x for x in size_d]
dp_u = [1] * (N + 1)

for v in order[1:]:
    p = parent[v]
    x = dp_d[p]
    x *= dp_u[p]
    x *= fact_inv[size_d[p]]
    x *= fact[size_d[v] + 1]
    x *= pow(dp_d[v], MOD - 2, MOD)
    x *= fact[size_u[v]]
    x *= fact_inv[size_u[p] + 1]
    dp_u[v] = x % MOD

for xd, xu, sd, su in zip(dp_d[1:], dp_u[1:], size_d[1:], size_u[1:]):
    su += 1
    x = xd * xu * fact[sd + su] * fact_inv[sd] * fact_inv[su] % MOD
    print(x)
