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
// Problem : F - Distributing Integers
// Contest : AtCoder Beginner Contest 160
// URL : https://atcoder.jp/contests/abc160/tasks/abc160_f
// Memory Limit : 1024 MB
// Time Limit : 3000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

#include <iostream>
#include <vector>
#define MOD 1000000007
using namespace std;
vector<int> e[200005];
long long siz[200005],f[200005],n;
long long fpow(long long x,long long y)
{
 long long ans=1;
 while(y)
 {
  if(y&1)ans=ans*x%MOD;
  x=x*x%MOD;
  y>>=1;
 }
 return ans;
}
void dfs1(int u,int fa)
{
 siz[u]=1;
 for(auto v:e[u])
  if(v!=fa)
  {
   dfs1(v,u);
   siz[u]+=siz[v];
  }
 f[1]=f[1]*siz[u]%MOD;
}
void dfs2(int u,int fa)
{
 for(auto v:e[u])
  if(v!=fa)
  {
   f[v]=f[u]*(n-siz[v])%MOD*fpow(siz[v],MOD-2)%MOD;
   dfs2(v,u);
  }
}
int main()
{
 cin>>n;
 for(int i=1;i<n;i++)
 {
  int u,v;
  cin>>u>>v;
  e[u].push_back(v);
  e[v].push_back(u);
 }
 f[1]=1;
 dfs1(1,0);
 dfs2(1,0);
 long long ans=1;
 for(int i=1;i<=n;i++)
  ans=ans*i%MOD;
 for(int i=1;i<=n;i++)
  cout<<ans*fpow(f[i],MOD-2)%MOD<<endl;
 return 0;
}