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;
template<class t> inline t read(t &x){
	char c=getchar();bool f=0;x=0;
	while(!isdigit(c)) f|=c=='-',c=getchar();
	while(isdigit(c)) x=(x<<1)+(x<<3)+(c^48),c=getchar();
	if(f) x=-x;return x;
}
template<class t> inline void write(t x){
	if(x<0) putchar('-'),write(-x);
	else{if(x>9) write(x/10);putchar('0'+x%10);}
}

#define int long long

const int mod=1e9+7,N=2e5+5;
int en,h[N],fac[N],inv[N],g[N],f[N],n,sz[N];

struct edge{
	int n,v;
}e[N<<1];

void add(int x,int y){
	e[++en]=(edge){h[x],y};
	h[x]=en;
}

int fpow(int x,int y){
	int res=1;
	for(;y;y>>=1,x=x*x%mod) if(y&1) res=res*x%mod;
	return res;
}

int C(int x,int y){
	return fac[x]*inv[y]%mod*inv[x-y]%mod;
}

void dfs1(int x,int fa){
	f[x]=1;
	for(int i=h[x];i;i=e[i].n){
		int y=e[i].v;
		if(y==fa) continue;
		dfs1(y,x);
		sz[x]+=sz[y];
		f[x]=f[x]*f[y]%mod*C(sz[x],sz[y])%mod;
	}
	sz[x]++;
}

void dfs2(int x,int fa){
	for(int i=h[x];i;i=e[i].n){
		int y=e[i].v;
		if(y==fa) continue;
		g[y]=f[y]*g[x]%mod*fpow(f[y]*C(n-1,sz[y])%mod,mod-2)%mod*C(n-1,n-sz[y])%mod;
		dfs2(y,x);
	}
}

signed main(){
	read(n);
	for(int i=1,x,y;i<n;i++){
		read(x);read(y);
		add(x,y);add(y,x);
	}
	fac[0]=1;
	for(int i=1;i<=n;i++) fac[i]=fac[i-1]*i%mod;
	inv[n]=fpow(fac[n],mod-2);
	for(int i=n-1;i>=0;i--) inv[i]=inv[i+1]*(i+1)%mod;
	dfs1(1,0);
	g[1]=f[1];
	dfs2(1,0);
	for(int i=1;i<=n;i++) write(g[i]),puts("");
}