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
// Etavioxy
#include<cstdio>
#include<cctype>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<functional>
#define ll long long
#define il inline
#define rep(i,s,t) for(register int i=(s);i<=(t);i++)
#define rev_rep(i,s,t) for(register int i=(s);i>=(t);i--)
#define each(i,u) for(int i=head[u];i;i=bow[i].nxt)
#define file(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)
#define pt(x) putchar(x)
using namespace std;
il int ci(){
	register char ch; int f=1;
	while(!isdigit(ch=getchar())) f=ch=='-'?-1:1;
	register int x=ch^'0';
	while(isdigit(ch=getchar())) x=(x*10)+(ch^'0');
	return f*x;
}

const int mod = 1e9+7;
enum{N=200023};

il ll qpow(ll a,ll b){
	ll ans= 1;
	for(; b; b>>=1,a=a*a%mod) if( b&1 ){
		ans= ans*a%mod;
	} return ans;
}
il ll inv(ll x){ return qpow(x,mod-2); }

struct Edge{ int nxt,to; } bow[N*2];
int head[N],tot_e;
il void add(int x,int y){
	tot_e++;
	bow[tot_e<<1]=(Edge){head[x],y};
	bow[tot_e<<1|1]=(Edge){head[y],x};
	head[y]=(head[x]=tot_e<<1)|1;
}

int fa[N],size[N];
void dfs1(int u){
	size[u] = 1;
	each(i,u){
		int v=bow[i].to;
		if( v==fa[u] ) continue;
		fa[v] = u;
		dfs1(v);
		size[u] += size[v];
	}
}
int n;
ll ans[N];
void dfs2(int u){
	if( fa[u] ){
		ans[u] = ans[fa[u]]*(n-size[u])%mod*inv(size[u])%mod;
	}
	each(i,u){
		int v=bow[i].to;
		if( v==fa[u] ) continue;
		dfs2(v);
	}
}


int main(){
	n=ci();
	rep(i,1,n-1) add(ci(),ci());
	dfs1(1);
	ans[1]=1;
	rep(i,1,n) ans[1]=ans[1]*size[i]%mod;
	dfs2(1);
	ll fac=1;
	rep(i,1,n) fac=fac*i%mod;
	rep(i,1,n) ans[i]=fac*inv(ans[i])%mod;
	rep(i,1,n) printf("%lld\n",(ans[i]+mod)%mod);
	return 0;
}
