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>
#define int long long
using namespace std;
const int N=2e5+5,MOD=1e9+7;
int n,u,v,tmp;
int size[N],f[N];
int cnt,head[N];
struct edge{int next,to;}e[N<<1];
 
inline int pow(int a,int n)
{
	int res=1;
	while (n)
	{
		if (n&1ll) res=res*a%MOD;
		a=a*a%MOD;
		n>>=1ll;
	}
	return res;
}
 
inline void add(int u,int v)
{
	cnt++;
	e[cnt].next=head[u];
	e[cnt].to=v;
	head[u]=cnt;
}
 
void dfs(int u,int fa)
{
	size[u]=1;
	for (register int i=head[u]; i; i=e[i].next)
	if (e[i].to!=fa)
	{
		dfs(e[i].to,u);
		size[u]+=size[e[i].to];
	}
	f[1]=f[1]*size[u]%MOD;
}
 
void dp(int u,int fa)
{
	for (register int i=head[u]; i; i=e[i].next)
	if (e[i].to!=fa)
	{
		f[e[i].to]=f[u]*pow(size[e[i].to],MOD-2)%MOD*(n-size[e[i].to])%MOD;
		dp(e[i].to,u);
	}
}
 
signed main(){
	scanf("%lld",&n);
	for (register int i=1; i<n; ++i) scanf("%lld%lld",&u,&v),add(u,v),add(v,u);
	f[1]=1; dfs(1,0);
	dp(1,0);
	tmp=1;
	for (register int i=1; i<=n; ++i) tmp=tmp*i%MOD;
	for (register int i=1; i<=n; ++i) 
	{	
		f[i]=tmp*pow(f[i],MOD-2)%MOD;
		printf("%lld\n",f[i]);
	}
return 0;	
}