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;
#define ll long long
#define N 200005
ll n,a,b,f[N],g=1,mod=1e9+7,fac[N],inv[N],size[N];
vector<int>v[N];
ll pows(ll x,ll y)
{
	ll ans=1;
	while(y)
	{
		if(y&1)ans=ans*x%mod;
		x=x*x%mod; 
		y>>=1;
	}
	return ans;
}
void pre(ll x,ll fa)
{
	size[x]=1;
	for(int i=0;i<v[x].size();i++)
	{
		int h=v[x][i];
		if(h==fa)continue;
		pre(h,x);
		size[x]+=size[h];
	}
	g=g*size[x]%mod;
}
void dfs(ll x,ll fa)
{
	for(int i=0;i<v[x].size();i++)
	{
		int h=v[x][i];
		if(h==fa)continue;
		f[h]=f[x]*(n-size[h])%mod*pows(size[h],mod-2)%mod;
		dfs(h,x);		
	}
}
int main()
{
	ios::sync_with_stdio(0);
	cin>>n;
	for(int i=1;i<n;i++)
	{
		cin>>a>>b;
		v[a].push_back(b);
		v[b].push_back(a);
	}
	ll nl=1;
	for(int i=1;i<=n;i++)nl=nl*i%mod;
	pre(1,1);
	f[1]=g;
	dfs(1,1);
	for(int i=1;i<=n;i++)
	printf("%lld\n",nl*pows(f[i],mod-2)%mod);
}