Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
1 2
1 3

Sample Output 1

2
1
1

The graph in this input is as follows:

For k=1, there are two ways in which we can write the numbers on the vertices, as follows:

Writing 1, 2, 3 on Vertex 1, 2, 3, respectively

Writing 1, 3, 2 on Vertex 1, 2, 3, respectively

Sample Input 2

2
1 2

Sample Output 2

1
1

The graph in this input is as follows:

Sample Input 3

5
1 2
2 3
3 4
3 5

Sample Output 3

2
8
12
3
3

The graph in this input is as follows:

Sample Input 4

8
1 2
2 3
3 4
3 5
3 6
6 7
6 8

Sample Output 4

40
280
840
120
120
504
72
72

The graph in this input is as follows:

Solution in Python (3.4.3) (judge verdict: Time Limit Exceeded)

class ReRooting:
    def __init__(self, size, root, identDp=0, identCum=0):
        """
        identDp: 木DP用 lift(a, identDp) = a
        identCum: 累積用 merge(a, identCum) = a
        """
        self.size = size
        self.edges = [[] for _ in range(self.size + 1)]
        self.root = root
        self.identDp = identDp
        self.identCum = identCum

    def addEdge(self, fr, to):
        self.edges[fr].append(to)
        self.edges[to].append(fr)

    # 根を固定したときの木DP
    def lift(self, top, bottom):
        return (((top[0] * bottom[0]) % MOD * comb.ncr(top[1] + bottom[1] + 1, top[1])) % MOD, top[1] + bottom[1] + 1)

    # 子の累積
    def merge(self, cum, a):
        return (((cum[0] * a[0]) % MOD * comb.ncr(cum[1] + a[1], cum[1])) % MOD, cum[1] + a[1])

    def sol(self):
        # 普通に木DPをする
        # 並行して各頂点につき、子の値の累積liftを左右から求めておく
        # その後根から順番に、親からの寄与を求めていく(fromParent)
        N = self.size
        edges = self.edges

        st = [self.root]
        parent = [-1] * (N + 1)
        child = [[] for _ in range(N + 1)]
        order = []
        while st:
            v = st.pop()
            order.append(v)
            for to in edges[v]:
                if parent[v] == to:
                    continue
                parent[to] = v
                child[v].append(to)
                st.append(to)

        dp = [self.identDp] * (N + 1)
        left = [self.identCum] * (N + 1)
        right = [self.identCum] * (N + 1)
        for v in order[::-1]:
            tmp = self.identCum
            for ch in child[v]:
                left[ch] = tmp
                tmp = self.lift(tmp, dp[ch])
            tmp = self.identCum
            for ch in reversed(child[v]):
                right[ch] = tmp
                tmp = self.lift(tmp, dp[ch])
            dp[v] = tmp

        fromParent = [self.identDp] * (N + 1)
        for v in order:
            if v == self.root:
                continue
            fromParent[v] = self.lift(self.merge(left[v], right[v]), fromParent[parent[v]])
            dp[v] = self.lift(dp[v], fromParent[v])

        return dp[:N]

class Combination:
    def __init__(self, size, MOD=10**9 + 7):
        self.size = size + 2
        self.MOD = MOD
        self.fact = [1, 1] + [0] * size
        self.factInv = [1, 1] + [0] * size
        self.inv = [0, 1] + [0] * size

        for i in range(2, self.size):
            self.fact[i] = self.fact[i - 1] * i % self.MOD
            self.inv[i] = -self.inv[self.MOD % i] * (self.MOD // i) % self.MOD
            self.factInv[i] = self.factInv[i - 1] * self.inv[i] % self.MOD

    def npr(self, n, r):
        if n < r or n < 0 or r < 0:
            return 0
        return self.fact[n] * self.factInv[n - r] % self.MOD

    def ncr(self, n, r):
        if n < r or n < 0 or r < 0:
            return 0
        return self.fact[n] * (self.factInv[r] * self.factInv[n - r] % self.MOD) % self.MOD

    def nhr(self, n, r):  # 重複組合せ
        return self.ncr(n + r - 1, n - 1)

    def factN(self, n):
        if n < 0:
            return 0
        return self.fact[n]

N = int(input())
MOD = 10**9 + 7
comb = Combination(2 * 10**5 + 100)

tree = ReRooting(N, 0, (1, -1), (1, 0))

for _ in range(N - 1):
    fr, to = map(int, input().split())
    tree.addEdge(fr - 1, to - 1)

ans = tree.sol()
print(*[a for a, b in ans], sep='\n')