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 <cstdio>
#include <algorithm>
#define LL long long
#define mod 1000000007
#define maxn 200010
using namespace std;
LL fact[maxn],inv[maxn];
int n,head[maxn],cnt,sz[maxn];
LL pre[maxn],ans[maxn],tp[maxn];
struct node{
	int to,next;
}e[maxn<<1];
void add_edge(int u,int v){
	cnt++,e[cnt].to=v,e[cnt].next=head[u],head[u]=cnt;
}
LL quick_pow(LL a,LL b){
	LL ans=1;
	while(b){
		if(b&1)ans=ans*a%mod;
		a=a*a%mod;
		b>>=1;
	}
	return ans;
}
void init(){
	int i,u,v;
	scanf("%d",&n);
	for(i=1;i<n;i++){
		scanf("%d%d",&u,&v);
		add_edge(u,v),add_edge(v,u);
	}
	fact[0]=1;
	for(i=1;i<=n;i++)fact[i]=fact[i-1]*i%mod;
	inv[n]=quick_pow(fact[n],mod-2);
	for(i=n-1;i>=0;i--)inv[i]=inv[i+1]*(i+1)%mod;
}
LL C(int a,int b){
	return fact[a]*inv[a-b]%mod*inv[b]%mod;
}
void dfs1(int u,int fa){
	int b,v;
	sz[u]=1,pre[u]=1;
	for(b=head[u];b;b=e[b].next){
		v=e[b].to;
		if(v==fa)continue;
		dfs1(v,u);
		sz[u]+=sz[v];
		pre[u]=pre[v]*pre[u]%mod*C(sz[u]-1,sz[v])%mod;
	}
}
void dfs2(int u,int fa){
	int b,v;
	ans[u]=tp[u]*pre[u]%mod*C(n-1,n-sz[u])%mod;
	for(b=head[u];b;b=e[b].next){
		v=e[b].to;
		if(v==fa)continue;
		tp[v]=ans[u]*quick_pow(pre[v]*C(n-1,sz[v])%mod,mod-2)%mod;
		dfs2(v,u);
	}
}
int main(){
	init();
	dfs1(1,0);
	tp[1]=1;
	dfs2(1,0);
	for(int i=1;i<=n;i++)printf("%lld\n",ans[i]);
	return 0;
}