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;
long long MOD=1000000007;
int n,uu,vv;
vector<int> ga[200100];
vector<int> ch[200100];
long long dp1[200100],dp2[200100];
long long inv[200100],fac[200100],ifac[200100];
int zsize[200100];
long long C(int x,int y){
	return fac[x]*ifac[y]%MOD*ifac[x-y]%MOD;
}
long long qpow(long long a,long long b){
	long long res=1LL;
	while(b){
		if(b&1)
			res=(res*a)%MOD;
		a=(a*a)%MOD;
		b>>=1;
	}
	return res%MOD;
}
void dp1_solve(int u,int p){
	zsize[u]=0;
	dp1[u]=1;
	for(int v:ga[u]){
		if(v==p)
			continue;
		ch[u].push_back(v);
		dp1_solve(v,u);
		zsize[u]+=zsize[v];
		dp1[u]=dp1[u]*C(zsize[u],zsize[v])%MOD*dp1[v]%MOD;
	}
	zsize[u]++;
}
void dp2_solve(int u){
	int tmp=zsize[u];
	for(int v:ch[u]){
		dp2[v]=dp2[u]*dp1[v]%MOD*qpow(dp1[v]*C(n-1,zsize[v])%MOD,MOD-2)%MOD*C(n-1,n-zsize[v])%MOD;
		dp2_solve(v);
	}
}
int main(){
	#ifdef FILIN
		#ifndef LOCAL
			freopen(FILIN,"r",stdin);
			freopen(FILOUT,"w",stdout);
		#endif
	#endif
	cin>>n;
	for(int i=1;i<n;i++){
		cin>>uu>>vv;
		ga[uu].push_back(vv);
		ga[vv].push_back(uu);
	}
	inv[1]=1LL;
	for(int i=2;i<=n;i++)
		inv[i]=(MOD-((MOD/i)*inv[MOD%i])%MOD);
	fac[0]=1LL;
	ifac[0]=1LL; 
	for(int i=1;i<=n;i++){
		fac[i]=(fac[i-1]*(long long)(i))%MOD;
		ifac[i]=(ifac[i-1]*inv[i])%MOD;
	}
	dp1_solve(1,-1);
	dp2[1]=dp1[1];
	dp2_solve(1);
	for(int i=1;i<=n;i++)
		cout<<(dp2[i])<<endl;
	return 0;
}
