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

def COMinit(n, MOD):
    fact = [1, 1]
    fact_inv = [1, 1]
    inv = [0, 1]
    for i in range(2, n + 1):
        fact.append((fact[-1] * i) % MOD)
        inv.append((-inv[MOD % i] * (MOD // i)) % MOD)
        fact_inv.append((fact_inv[-1] * inv[i]) % MOD)
    return fact, fact_inv

MOD = 10**9+7
def resolve():
    def dfs1(r_topo, par):
        for idx in reversed(r_topo):
            for to in G[idx]:
                if to == par[idx]:
                    continue


    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)

    fact, fact_inv = COMinit(N, MOD)

    topo = []
    P = [-1] * N
    node = [0]
    while node:
        v = node.pop()
        topo.append(v)
        for to in G[v]:
            if to == P[v]:
                continue
            P[to] = v
            node.append(to)

    # 葉のサイズを格納
    size = [1]*N
    for e in reversed(topo):
        if e == 0:
            break
        size[P[e]] += size[e]

    dp1 = [1]*N
    for e in reversed(topo):
        dp1[e] = dp1[e] * fact[size[e] - 1] % MOD
        if e == 0:
            break
        dp1[P[e]] = (dp1[P[e]] * dp1[e] % MOD) * fact_inv[size[e]] % MOD

    dp2 = [0] * N
    dp2[0] = dp1[0]
    for e in topo[1:]:
        p = P[e]
        tmp = (dp2[p] * fact[N - 1 - size[e]] % MOD) * fact[size[e]] % MOD
        red = (tmp * fact_inv[N - 1] % MOD) * pow(dp1[e], MOD - 2, MOD) % MOD

        size_red = N - size[e]
        tmp = (dp1[e] * red % MOD) * fact[N - 1] % MOD
        dp2[e] = (tmp * fact_inv[size[e] - 1] % MOD) * fact_inv[size_red] % MOD

    print(*dp2, sep="\n")

if __name__ == "__main__":
    resolve()
