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**9)

def solve():
    MOD = 10**9 + 7

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

    def getInvs(n, MOD):
        invs = [1] * (n+1)
        for x in range(2, n+1):
            invs[x] = (-(MOD//x) * invs[MOD%x]) % MOD
        return invs
    invs = getInvs(N, MOD)

    dp = [[] for _ in range(N)]
    outdegs = [0] * N
    sizes = [0] * N
    iPars = [-1] * N
    def dfsDP(v, vPar):
        outdeg = outdegs[v] = len(adjL[v])
        dp[v] = [0] * outdeg
        res = 1
        sizes[v] = 1
        for i, v2 in enumerate(adjL[v]):
            if v2 == vPar:
                iPars[v] = i
                continue
            dp[v][i] = dfsDP(v2, v)
            res = res * dp[v][i] % MOD
            sizes[v] += sizes[v2]
        return res * invs[sizes[v]] % MOD

    anss = [0] * N
    def dfsAns(v, vPar, resPar):
        if vPar != -1:
            dp[v][iPars[v]] = resPar

        outdeg = outdegs[v]
        dpL = [1] * (outdeg+1)
        for i in range(outdeg):
            v2 = adjL[v][i]
            dpL[i+1] = dpL[i] * dp[v][i] % MOD
        dpR = [1] * (outdeg+1)
        for i in reversed(range(outdeg)):
            v2 = adjL[v][i]
            dpR[i] = dpR[i+1] * dp[v][i] % MOD
        anss[v] = dpL[-1] * invs[N]

        for i, v2 in enumerate(adjL[v]):
            if v2 == vPar: continue
            dfsAns(v2, v, dpL[i] * dpR[i+1] * invs[N-sizes[v2]] % MOD)


    dfsDP(0, -1)
    dfsAns(0, -1, 0)

    def getFact(n, MOD):
        fact = 1
        for i in range(1, n+1):
            fact = (fact * i) % MOD
        return fact
    factN = getFact(N, MOD)

    for v in range(N):
        anss[v] = anss[v] * factN % MOD
    print('\n'.join(map(str, anss)))


solve()
