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
n = int(input()); arr = [list(map(int,input().split())) for _ in range(n-1)]
graph = [[] for _ in range(n)]
for a,b in arr:
    graph[a-1].append(b-1); graph[b-1].append(a-1)


mod = 10**9+7
fac = [1]*(n+1)
inv = [1]*(n+1)
ifac = [1]*(n+1)
for i in range(2,n+1):
    fac[i] = fac[i-1] * i % mod
    inv[i] = (- (mod//i)*inv[(mod%i)]) % mod
    ifac[i] = ifac[i-1] * inv[i] % mod

parents = [-1]*n
children = [[] for _ in range(n)]
orders = []
stack = [0]
flags = [True] * n; flags[0] = False
while stack:
    tmp = stack.pop()
    orders.append(tmp)
    for node in graph[tmp]:
        if flags[node]:
            flags[node] = False
            children[tmp].append(node)
            stack.append(node)
            parents[node] = tmp

size = [1] * (n+1)
dp = [1] * (n+1)    
for child in orders[::-1]:
    parent = parents[child]
    size[parent] += size[child]
    dp[child] = dp[child] * fac[size[child]-1] % mod
    dp[parent] = dp[parent] * dp[child] * ifac[size[child]] % mod
_ = size.pop();_ = dp.pop()

size2 = [n-i+1 for i in size]
dp2 = [1] * n; dp2[0] = dp[0]
for parent in orders[1:]:
    child = parents[parent]
    dp2[parent] = dp2[parent] * size[parent] * dp2[child] * inv[size2[parent]-1] % mod
    
for i in range(n):
    print(dp2[i])