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>
using namespace std;
typedef long long ll;
#define p(a) printf("%lld\n",a);
ll n;
const ll mod=1e9+7;
vector<ll> g[200005];
bool used[200005]={0};
ll size[200005]={0};
ll njie[200005]={0};
ll nfan[200005]={0};
ll sum[200005]={0};
ll a[200005]={0};
ll qpow(ll a,ll b)
{
	if(b==0) return 1;
	else if(b==1) return a;
	ll c=qpow(a,b/2);
	if(b%2==0) return c*c%mod;
	else return c*c%mod*a%mod;
}
void dfs1(ll u)
{	//printf("%lld ",u);
	used[u]=true;
	for(ll i=0;i<g[u].size();i++)
	{
		ll v=g[u][i];
		if(used[v]) continue;
		dfs1(v);
		size[u]+=size[v];
		a[u]=a[u]*a[v]%mod*nfan[size[v]]%mod;
	}
	a[u]=a[u]*njie[size[u]-1]%mod;
}
void change(ll u,ll v)
{
	a[u]=a[u]*nfan[size[u]-1]%mod;
	a[u]=a[u]*njie[size[u]-size[v]-1]%mod;
	a[u]=a[u]*njie[size[v]]%mod;
	a[u]=a[u]*qpow(a[v],mod-2)%mod;
	size[u]=size[u]-size[v];
	a[v]=a[v]*nfan[size[v]-1]%mod;
	size[v]=size[v]+size[u];
	a[v]=a[v]*njie[size[v]-1]%mod;
	a[v]=a[v]*a[u]%mod;
	a[v]=a[v]*nfan[size[u]]%mod;
}
void dfs2(ll u)
{
	sum[u]=a[u];
	used[u]=true;
	for(ll i=0;i<g[u].size();i++)
	{
		ll v=g[u][i];
		if(!used[v])
		{
			change(u,v);
			dfs2(v);
			change(v,u);
		}
	}
}
int main()
{
	scanf("%lld",&n);
	njie[0]=1;
	for(ll i=1;i<=n;i++) njie[i]=njie[i-1]*i%mod,a[i]=1,size[i]=1;
	nfan[n]=qpow(njie[n],mod-2);
	for(ll i=n-1;i>0;i--) nfan[i]=nfan[i+1]*(i+1)%mod;
	nfan[0]=nfan[1];
	for(ll i=0;i<n-1;i++)
	{
		ll u,v;
		scanf("%lld%lld",&u,&v);
		g[u].push_back(v);
		g[v].push_back(u);
	}
	dfs1(1);
	memset(used,0,sizeof(used));
	dfs2(1);
	for(ll i=1;i<=n;i++)
	{
		printf("%lld ",sum[i]);
	}
}