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 ll long long
using namespace std;
const ll p=1e9+7;
ll n,tot,e[500010],hd[500010],nt[500010],sz[500010],ans,a[500010];
void build(int x,int y){
	tot++;
	e[tot]=y;
	nt[tot]=hd[x];hd[x]=tot;
}
void dfs(int x,int fa){
	int i;
	sz[x]=1;
	for(i=hd[x];i;i=nt[i]){
		if(e[i]==fa)continue;
		dfs(e[i],x);
		sz[x]+=sz[e[i]];
	}
	ans=ans*sz[x]%p;
}
ll ksm(ll x,ll y){
	ll xlh=1;
	while(y){
		if(y&1)xlh=xlh*x%p;
		x=x*x%p;
		y/=2;
	}
	return xlh;
}
void get(int x,int fa,int la){
	a[x]=ans*la%p*ksm(sz[x],p-2)%p*n%p;
	int i;
	for(i=hd[x];i;i=nt[i]){
		if(e[i]==fa)continue;
		get(e[i],x,la*ksm(sz[x],p-2)%p*(n-sz[e[i]])%p);
	}
}
int main(){
	ll i,x,y,jc;
	scanf("%lld",&n);
	ans=1;jc=n;
	for(i=1;i<n;i++){
		scanf("%lld%lld",&x,&y);
		build(x,y);build(y,x);
		jc=jc*i%p;
	}
	dfs(1,0);
	get(1,0,1);
	for(i=1;i<=n;i++)printf("%lld\n",jc*ksm(a[i],p-2)%p);
}