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
input = sys.stdin.readline
sys.setrecursionlimit(10**6)

N = int(input())
G = [[] for _ in range(N)]
for _ in range(N-1):
    a, b = map(lambda x : int(x)-1, input().split())
    G[a].append(b)
    G[b].append(a)

mod = 10**9+7
fact = [1] * N
fact_inv = [1] * N
for i in range(1, N):
    fact[i] = i * fact[i-1] % mod
inv = lambda x : pow(x, mod-2, mod)
fact_inv[N-1] = inv(fact[N-1])
for i in range(1, N)[::-1]:
    fact_inv[i-1] = i * fact_inv[i] % mod

stack = [0]
parent = [-1] * N
order = []
while stack:
    v = stack.pop()
    order.append(v)
    for c in G[v]:
        if c != parent[v]:
            stack.append(c)
            parent[c] = v

dp = [1] * N
size = [1] * N
for v in order[::-1]:
    dp[v] = dp[v] * fact[size[v]-1] % mod
    p = parent[v]
    if p == -1:
        break
    size[p] += size[v]
    dp[p] = dp[p] * dp[v] * fact_inv[size[v]] % mod

ans = [0] * N
ans[0] = dp[0]
for v in order[1:]:
    p = parent[v]
    ans[v] = ans[p] * size[v] * inv(N-size[v]) % mod

print(*ans, sep='\n')