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<bits/stdc++.h>

const int N = 2 * 1e5 + 10;
const int mod = 1e9 + 7;
using namespace std;

int n;
int quick_pow(int a, int b)
{
	int res = 1;
	while(b)
	{
		if(b&1)
		{
			res = 1ll * res * a % mod;
		}
	   	b >>= 1;
	   	a = 1ll*a*a%mod;
	}
	return res;
}
struct edge
{
	int y;
	int next;
}ed[2*N];
int en = 0;
int head[N];
void add_edge(int x, int y)
{
	en++;
	ed[en].y = y;
	ed[en].next = head[x];
	head[x] = en;
}
int f[N];
void dfs(int u, int fa)
{
	for(int p = head[u]; p != 0; p = ed[p].next)
	{
		int v = ed[p].y;
		if(v == fa)continue; 
			dfs(v, u);
			f[u] += f[v];
	}
	f[u]++;
}
int ans[N];
void move(int u, int fa)
{
	for(int p = head[u]; p != 0; p = ed[p].next)
	{
		int v = ed[p].y;
		if(v == fa)
		{
			continue;
		}
		ans[v] =  1ll * ans[u] * quick_pow(f[v], mod-2) % mod * (n - f[v]) % mod;
		move(v, u);
	}

}
int main()
{
	cin >> n;
	for(int i = 1; i <= n-1; i++)
	{
		int a, b;
		cin >> a >> b;
		add_edge(a, b);
		add_edge(b, a);
	}
	dfs(1, 0);
	ans[1] = 1;
	for(int i = 1; i <= n; i++)
	{
		ans[1] = 1ll * ans[1] * f[i] % mod;
	}
	move(1, 0);
	int t = 1;
	for(int i = 1; i <= n; i++)
	{
		t = 1ll * i * t % mod;
	}
	for(int i = 1; i <= n; i++)
	{
		printf("%d\n", 1ll * t * quick_pow(ans[i], mod-2) % mod);
	}
	return 0;
}
