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"
#include<iostream>
#include<vector>
using namespace std;
typedef long long ll;
// using pAr = product_as_root;
const ll mod = 1e9+7;
ll n;
ll sn[200005];
ll pAr[200005];
vector<ll> G[200005];
ll qpow(ll x,ll n,ll mod)
{
	ll res =1;
	while(n>0)
	{
		if(n&1) res=res*x%mod;
		x=x*x%mod;
		n>>=1;
	}
	return res;
}
void dfs_sn(ll u,ll v){
	for(auto i:G[u]){
		if(i!=v){
			dfs_sn(i,u); sn[u]+=sn[i];
		}
	}
	sn[u]++;
}
void dfs_cr(ll u,ll v){
	for(auto i:G[u]){
		if(i!=v){
			pAr[i]=pAr[u]*qpow(sn[i],mod-2,mod)%mod*(n-sn[i])%mod;
			dfs_cr(i,u);
		}
	}
}
int main(){
	//freopen("input.txt","r",stdin);
	cin>>n;
	ll t1,t2;
	for(ll i=0;i<n-1;i++) {
		cin>>t1>>t2;
		G[t1].push_back(t2);
		G[t2].push_back(t1);
	}
	dfs_sn(1,0);
	//init
	pAr[1]=1;
	for(ll i=1;i<=n;i++) pAr[1]=pAr[1]*sn[i]%mod;
	dfs_cr(1,0);
	//factorial
	ll fac=1;
	for(ll i=2;i<=n;i++) fac=fac*i%mod;
	for(ll i=1;i<=n;i++){
		cout<<fac*qpow(pAr[i],mod-2,mod)%mod<<endl;
	}
	//for(int i=1;i<=8;i++) cout<<pAr[i]<<" ";
	return 0;
}


