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
/*
luogu AT4826
*/
#include <bits/stdc++.h>
using namespace std;

#define int long long

const int M = 1e9+7ll;

int n;
int siz[200005],f[200005];
int ee,h[200005],nex[200005<<1],to[200005<<1];

int ksm(int x,int w)
{
	int s=1;
	while(w)
	{
		if(w&1)
			s = s*x%M;
		x = x*x%M;
		w >>= 1;
	}
	return s%M;
}

void addedge(int x,int y)
{
	nex[++ee] = h[x];
	to[ee] = y;
	h[x] = ee;
}

void getsiz(int x,int pre)
{
	siz[x] = 1;
	for(int i=h[x];i;i=nex[i])
		if(to[i]!=pre)
		{
			getsiz(to[i],x);
			siz[x] += siz[to[i]];
		}
	f[1] = f[1]*siz[x]%M;
}

void dfs(int x,int pre)
{
	for(int i=h[x];i;i=nex[i])
		if(to[i]!=pre)
		{
			f[to[i]] = f[x]*(n-siz[to[i]])%M*ksm(siz[to[i]],M-2)%M;
			dfs(to[i],x);
		}
}

signed main()
{
	cin>>n;
	for(int i=1,x,y;i<n&&cin>>x>>y;i++)
		addedge(x,y), addedge(y,x);
	f[1] = 1;
	getsiz(1,0);
	dfs(1,0);
	int s=1;
	for(int i=1;i<=n;i++)
		s = s*i%M;
	for(int i=1;i<=n;i++)
		cout<<s*ksm(f[i],M-2)%M<<endl;
	
	return 0;
}