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<iostream>
#include<vector>
using namespace std;
const long mod=1e9+7;
long power(long a,long b){return b?power(a*a%mod,b/2)*(b%2?a:1)%mod:1;}
long F[2<<17],I[2<<17];
int N;
vector<int>G[2<<17];
pair<int,long>dp1[2<<17];
long ans[2<<17];
pair<int,long>dfs1(int u,int p)
{
	pair<int,long>ret=make_pair(0,1);
	for(int v:G[u])
	{
		if(v==p)continue;
		pair<int,long>q=dfs1(v,u);
		ret.first+=q.first;
		(ret.second*=q.second*I[q.first]%mod)%=mod;
	}
	(ret.second*=F[ret.first++])%=mod;
	dp1[u]=ret;
	return ret;
}
void dfs2(int u,int p,pair<int,long>pr)
{
	pair<int,long>ret=dp1[u];
	ans[u]=ret.second*I[ret.first-1]%mod;
	(ans[u]*=pr.second*I[pr.first]%mod)%=mod;
	for(int v:G[u])
	{
		if(v==p)continue;
		pair<int,long>nxt;
		nxt.first=N-dp1[v].first;
		nxt.second=ans[u]*power(dp1[v].second,mod-2)%mod*F[dp1[v].first]%mod*F[nxt.first-1]%mod;
		dfs2(v,u,nxt);
	}
	(ans[u]*=F[ret.first-1+pr.first])%=mod;
}
int main()
{
	F[0]=1;
	for(int i=1;i<2<<17;i++)F[i]=F[i-1]*i%mod;
	I[(2<<17)-1]=power(F[(2<<17)-1],mod-2);
	for(int i=(2<<17)-1;i--;)I[i]=I[i+1]*(i+1)%mod;
	cin>>N;
	for(int i=1;i<N;i++)
	{
		int u,v;cin>>u>>v;
		u--,v--;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	dfs1(0,-1);
	dfs2(0,-1,make_pair(0,1));
	for(int i=0;i<N;i++)
	{
		cout<<ans[i]<<endl;
	}
}
