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
#include <iostream>
#include <vector>
using namespace std;

int n, a, b;
long long g = 1;
int p[200001], s[200001];
const long long M = 1000000007;
vector<int> e[200001];
long long f[200001];

long long pw(long long a, long long b) {
	long long c = 1, d = a;

	while (b) {
		if (b & 1)
			c = c * d % M;

		d = d * d % M;
		b >>= 1;
	}

	return c;
}

void pd(int id) {
	s[id] = 1;
	f[id] = 1;

	for (int i : e[id]) {
		if (!s[i]) {
			pd(i);
			s[id] += s[i];
			p[i] = id;
			f[id] = f[id] * f[i] % M;
		}
	}
	f[id] = f[id] * s[id] % M;
}

void dfs(int id) {
	for (int i : e[id]) {
		if (i != p[id]) {
			f[i] = f[id] * (n - s[i]) % M * pw(s[i], M - 2) % M;
			dfs(i);
		}
	}
}

int main()
{
	cin >> n;

	for (int i = 1; i < n; i++) {
		cin >> a >> b;
		e[a].push_back(b);
		e[b].push_back(a);
	}

	for (int i = 1; i <= n; i++)
		g = g * i % M;

	pd(1);
	dfs(1);

	for (int i = 1; i <= n; i++)
		cout << g * pw(f[i], M - 2) % M << endl;
}
