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>
#define pb push_back
#define mod int(1e9+7)
#define N int(2e5+100)
#define ll long long int
#define pp pair<int,int>
#define pl pair<ll,ll>
#define fi first
#define si second
using namespace std;
ll n,u,v,par[N],size[N],ans[N];
ll dp[N],fact[N];
vector<ll>g[N];
ll power(ll x,ll n){
	if(n==0){
		return 1ll;
	}
	ll y=power(x,n/2);
	y=(y*y)%mod;
	if(n%2){
		return (x*y)%mod;
	}
	return y;
}
void pre(){
	fact[0]=1;
	for(ll i=1;i<N;i++){
		fact[i]=(i*fact[i-1])%mod;
	}
}
void dfs(ll s,ll p){
	par[s]=p;
	size[s]=1;
	for(ll i=0;i<g[s].size();i++){
		if(g[s][i]!=p){
			dfs(g[s][i],s);
			size[s]+=size[g[s][i]];
		}
	}
	return;
}
void dfs1(ll s,ll p){
	ll d=1ll;
	dp[s]=fact[size[s]-1];
	for(ll i=0;i<g[s].size();i++){
		if(g[s][i]!=p){
			dfs1(g[s][i],s);
			d=(d*fact[size[g[s][i]]])%mod;
			dp[s]=(dp[s]*dp[g[s][i]])%mod;
		}
	}
	dp[s]=(dp[s]*power(d,mod-2))%mod;
}
ll reRoot(ll s,ll p){
	if(p==-1 ) {
		return dp[s];
	}
	if(ans[s]){
		return ans[s];
	}
	ll dpp=reRoot(p,par[p]);
	ll dps=dp[s],szp=n-size[s],d;
	d=(fact[n-1]*dps)%mod;
	dpp=(dpp*power(d,mod-2))%mod;
	dpp=(dpp*fact[size[s]])%mod;
	dpp=(dpp*fact[szp-1])%mod;
	d=(fact[size[s]-1]*fact[szp])%mod;
	dps=(dps*power(d,mod-2))%mod;
	dps=(dps*dpp)%mod;
	dps=(dps*fact[n-1])%mod;
	ans[s]=dps;
	return ans[s];
}
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
	cin>>n;
	pre();
	for(ll i=1;i<n;i++){
		cin>>u>>v;
		g[v].pb(u);
		g[u].pb(v);
	}
	dfs(1,-1);
	dfs1(1,-1);
	ans[1]=dp[1];
	for(ll k=1;k<=n;k++){
		cout<<reRoot(k,par[k])<<endl;
	}
    return 0;
}

