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>
#define M 300005 
#define LL long long
using namespace std;
int A[M],B[M],C[M],sz[M],fa[M];
int P=1e9+7,n;
vector<int>edge[M];
LL F[M],dp[M],pre;
int fast(int x,int n){
	LL res=1,a=x;
	while(n){
		if(n&1)res=res*a%P;
		a=a*a%P;
		n>>=1;
	}
	return res;
}
void dfs(int x,int f){
	sz[x]=1;fa[x]=f;
	for(int i=0;i<edge[x].size();i++){
		int y=edge[x][i];
		if(y==f)continue;
		dfs(y,x);
		sz[x]+=sz[y];//儿子的数量  
	}
} 
void rdfs(int x,int f,LL pre){ 
	 //pre记录x的祖先节点在把x变成根是的前缀积 
	for(int i=0;i<edge[x].size();i++){
		int y=edge[x][i];
		if(y==f)continue; 
		//y变成根之后，y在原树的所有祖先节点的 子树大小都发生变化 
		LL p=pre*sz[x]%P*fast(n-sz[y],P-2)%P; 
		LL tmp=p*sz[y]%P*fast(n,P-2)%P; 
		dp[y]=dp[1]*tmp%P; 
		rdfs(y,x,p); 
	}
}
int main(){
	int a,b;
	cin>>n;
	F[0]=1;
	for(int i=1;i<=n;i++)F[i]=F[i-1]*i%P;
	for(int i=1;i<n;i++){
		scanf("%d %d",&a,&b);
		edge[a].push_back(b);
		edge[b].push_back(a);
	}
	dfs(1,0);
	dp[1]=F[n];
	for(int i=1;i<=n;i++)dp[1]=dp[1]*fast(sz[i],P-2)%P; 
	rdfs(1,0,1);
	for(int i=1;i<=n;i++)printf("%lld\n",dp[i]);
	return 0; 
} 