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
#pragma GCC optimize ("O3")
#pragma GCC optimize("unroll-loops")
#include"bits/stdc++.h"
using namespace std;
typedef long long ll;
#define int ll
const int mod=1e9+7;
int n;
const int N=2e5;
int f[N],inv[N];
vector<vector<int>> ad;
int sz[N];
int dp[N];
int ans[N];
int modp(int a,int b)
{
	if(b==0)return 1;
	int u=modp(a,b/2);
	u=(u*u)%mod;
	if(b&1)return (a*u)%mod;
	return u;
}
void dfs1(int node=0, int par=-1)
{
	sz[node]=1;dp[node]=1;
	for(int u:ad[node])
	{
		if(u!=par)
		{
			dfs1(u,node);
			sz[node]+=sz[u];
			dp[node]*=dp[u];
			dp[node]%=mod;
		}
	}
	dp[node]*=sz[node];
	dp[node]%=mod;
}
void dfs2(int u=0, int par=-1, int abovediv=1)
{
	ans[u]=f[n-1];
	ans[u]*=modp(abovediv,mod-2);ans[u]%=mod;
	ans[u]*=modp(dp[u],mod-2);ans[u]%=mod;ans[u]*=sz[u];ans[u]%=mod;
	
	for(auto v:ad[u])
	{
		if(v!=par)
		{
			int pass=dp[u]*abovediv;pass%=mod;pass*=(n-sz[v]);pass%=mod;pass*=modp(dp[v],mod-2);
			pass%=mod;pass*=modp(sz[u],mod-2);
			pass%=mod;
			dfs2(v,u,pass);
		}
	}
	
	
	
	
}
int32_t main()
{
	#ifdef ONLINE_JUDGE
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	#endif
	cin>>n;
	int u,v;
	f[0]=1;
	for(int i=1;i<N;i++)
	f[i]=(i*f[i-1])%mod;
	inv[N-1]=modp(f[N-1],mod-2);
	for(int i=N-2;i>=0;i--)inv[i]=((i+1)*inv[i+1])%mod;
	ad=vector<vector<int>>(n+1);
	for(int i=0;i<n-1;i++)
	{
		cin>>u>>v;
		u--;v--;
		ad[u].push_back(v);
		ad[v].push_back(u);
	}
	
	dfs1();
	dfs2();
	for(int i=0;i<n;i++)cout<<ans[i]<<'\n';
}