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 <stdio.h>
#include <algorithm>
#include <vector>
using namespace std;

const long long mod = 1000000007;
long long inv[200200] = { 0, 1 };

int n; vector<int> g[200200];

long long d[200200]; int s[200200];

void go1(int x, int l)
{
	d[x] = 1;
	s[x] = 1;
	for (auto y : g[x]) if (y != l){
		go1(y, x);
		d[x] = d[x] * d[y] % mod * inv[s[y]] % mod;
		s[x] += s[y];
	}
}

long long finv(long long x)
{
	long long r = 1;
	long long p = mod - 2;
	while (p){
		if (p & 1) r = r * x % mod;
		x = x * x % mod;
		p /= 2;
	}
	return r;
}

long long u[200200];

void go2(int x, int l)
{
	for (auto y : g[x]) if (y != l){
		if (x == 1) u[y] = d[x] % mod * finv(d[y]) % mod * s[y] % mod;
		else u[y] = u[x] * d[x] % mod * finv(d[y]) % mod * s[y] % mod * inv[n - s[x]] % mod;
		go2(y, x);
		d[y] = d[y] * u[y] % mod * inv[n - s[y]] % mod;
	}
}

int main()
{
	for (int i = 2; i < 200200; i++) inv[i] = (mod - mod / i) * inv[mod % i] % mod;

	scanf ("%d", &n);
	for (int i = 1; i < n; i++){
		int x, y; scanf ("%d %d", &x, &y);
		g[x].push_back(y);
		g[y].push_back(x);
	}

	go1(1, 0);
	go2(1, 0);

	long long f = 1;
	for (int i = 1; i < n; i++) f = f * i % mod;
	for (int i = 1; i <= n; i++) printf ("%lld\n", d[i] * f % mod);

	return 0;
}