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 <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <cmath>
using namespace std;
typedef long long ll;
const ll N=410000,mod=1e9+7;
ll n,m,T,cnt,a[N],fi[N],ne[N],la[N],siz[N],g[N],ans[N],x,y,z,k,aa,bb,cc;
ll qpow(ll x,ll p)
{
	ll ans=1,ret=x;
	for (;p;p/=2)
	{
		if (p%2==1) ans=(ans*ret)%mod;
		ret=(ret*ret)%mod;
	}
	return ans;
}
void add(ll x,ll y)
{
	a[++k]=y;
	if (!fi[x]) fi[x]=k; else ne[la[x]]=k;
	la[x]=k;
}
void dfs(int x,int fa)
{
	siz[x]=1;
	for (int i=fi[x];i;i=ne[i]) if (a[i]!=fa)
	{
		dfs(a[i],x);
		siz[x]+=siz[a[i]];
	}
}
void dfs2(int x,int fa)
{
	if (x!=1) 
	{
		ans[x]=ans[fa]*siz[x]%mod;
		ans[x]=ans[x]*qpow(n-siz[x],mod-2)%mod;
	}
	for (int i=fi[x];i;i=ne[i]) if (a[i]!=fa) dfs2(a[i],x);
	
}
int main() 
{
	scanf("%lld",&n);
	for (int i=1;i<n;i++)
	{
		scanf("%lld%lld",&x,&y);
		add(x,y); add(y,x);
	}
	dfs(1,0);
	ans[1]=1;
	for (ll i=1;i<=n;i++) 
	{
		ans[1]=(ans[1]*i)%mod;
		ans[1]=(ans[1]*qpow(siz[i],mod-2))%mod;
	}
	dfs2(1,0);
	for (int i=1;i<=n;i++) printf("%lld\n",ans[i]);
}