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
//love yjl forever
#include<bits/stdc++.h>
using namespace std;
#define ll long long 
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define fi first
#define se second
const ll mo=1e9+7,N=2e5+5;
int n,m;
ll sz[N],ps[N];
ll wow[N],ny[N];
ll ans[N],wai[N],dp[N];
vector<int> v[N];
ll ksm(ll x,ll y)
{
	if(y==0) return 1;
	ll tmp=ksm(x,y/2);
	
	if(y&1)
	{
		return ((tmp*tmp)%mo*x)%mo;
	}
	else 
		return (tmp*tmp)%mo;
}
void dfs(int x,int fa)
{
	ll s1=1,s2=1;
	ps[x]=1;
	for(int i=0;i<v[x].size();i++)
	{
		int u=v[x][i];
		if(u==fa) continue;
		dfs(u,x);
		sz[x]+=sz[u];
//		ps[x]=(ps[x]*sz[u])%mo;
		s1=(s1*ny[sz[u]])%mo;
		s2=(s2*ans[u])%mo;
	}
	sz[x]++;
	ps[x]=s1;
	ans[x]=((wow[sz[x]-1]*s1)%mo*s2)%mo;
}
void dfs2(int x,int fa)
{
	for(int i=0;i<v[x].size();i++)
	{
		int u=v[x][i];
		if(u==fa) continue;
		wai[u]=((((dp[x]*wow[(n-sz[u]-1)])%mo*ny[n-1])%mo*ksm(ans[u],mo-2))%mo*wow[sz[u]])%mo;
		dp[u]=((((ans[u]*ny[sz[u]-1])%mo*wow[n-1])%mo*wai[u])%mo*ny[n-sz[u]])%mo;
		dfs2(u,x);
	}
}
int main()
{
/*	freopen("","r",stdin);
	freopen("","w",stdout);*/
	int i;
	cin>>n;
	for(i=1;i<n;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		v[x].pb(y);
		v[y].pb(x);
	}
	wow[0]=1;ny[0]=1;
	for(i=1;i<=n;i++)
		wow[i]=(wow[i-1]*i)%mo,ny[i]=ksm(wow[i],mo-2)%mo;
	
//	for(i=1;i<n;i++)
//		cout<<wow[i]<<" ";
//	cout<<endl;
//	for(i=1;i<n;i++)
//		cout<<ny[i]<<" ";
//	cout<<endl;	
	for(i=1;i<=n;i++)
		ans[i]=1;

	dfs(1,1);

	dp[1]=ans[1];
	dfs2(1,1);

	for(i=1;i<=n;i++)
		cout<<dp[i]<<endl;
//	cout<<endl;
//	for(i=1;i<=n;i++)
//		cout<<sz[i]<<" ";
//	cout<<endl;
//	for(i=1;i<=n;i++)
//		cout<<ans[i]<<" ";
//	cout<<endl;
}