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;
typedef long long ll;
const int N=32e4,M=1e9+7;

vector<int>t[N];
bool u[N];
int s[N],n;
ll d[N],a[N],f[N];
ll qpow(ll a,ll x){
	ll ans=1;
	for(;x;x>>=1){
		if(x&1)ans=ans*a%M;
		a=a*a%M;
	}
	return ans;
}
ll C(int n,int m){
	return f[n]*qpow(f[m]*f[n-m]%M,M-2)%M;
}
ll dfs(int x){
	u[x]=d[x]=1;
	for(int y:t[x])if(!u[y]){
		s[x]+=dfs(y);
		d[x]=d[x]*C(s[x],s[y])%M*d[y]%M;
	}
	u[x]=0;
	return ++s[x];
}
void get_ans(int x,ll z){
	u[x]=1;
	a[x]=d[x]*C(n-1,n-s[x])%M*z%M;
	for(int y:t[x])if(!u[y])
		get_ans(y,a[x]*qpow(d[y]*C(n-1,s[y])%M,M-2)%M);
	u[x]=0;
}
int main(){
	ios::sync_with_stdio(0);
	cin>>n;
	f[0]=1;
	for(int i=1,u,v;i<n;i++){
		f[i]=f[i-1]*i%M;
		cin>>u>>v;
		t[u].push_back(v);
		t[v].push_back(u);
	}
	dfs(1);
	get_ans(1,1);
	for(int i=1;i<=n;i++)cout<<a[i]<<'\n';
	return 0;
}