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;
const ll N=2e5+5,mod=1e9+7;
vector<ll>g[N];
ll a[N],n,sz[N],tmp=1; //a[i]表示以结点i为开始点对排列的影响。 
ll ksm(ll a,ll n){ //ksm(快速幂） 
	ll ans=1;
	while(n){
		if(n&1) ans=ans*a%mod;
		a=a*a%mod;
		n>>=1;
	}
	return ans;
}
void dfs(int u,int fa){
	sz[u]=1;
	for(auto v:g[u]){
		if(v!=fa){
			dfs(v,u);
			sz[u]+=sz[v];
		}
	}
	tmp=tmp*sz[u]%mod;//sz[ i ]的连乘 
}
void fun(int u,int fa){ //状态转移方程. 
	 for(auto v:g[u]){
	 	 if(v!=fa){
	 	 	 a[v]=a[u]*ksm(sz[v],mod-2)%mod*(n-sz[v])%mod;
			 fun(v,u);	
		  }
	 }
}
int main(){
	cin>>n; 
	ll jc=1;//jc ( 阶乘 ) 
	for(ll i=2,u,v;i<=n;i++){
		 cin>>u>>v;
		 g[u].push_back(v),g[v].push_back(u);
		 jc=jc*i%mod; 
	}
	dfs(1,0);
	a[1]=tmp;
	fun(1,0); 
		for(int i=1;i<=n;i++)
			printf("%lld\n",jc*ksm(a[i],mod-2)%mod);
	return 0;
} 