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 <cstdio>
#include <iostream>
#include <cmath>
#include <set>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <cstring>
#include <string>
using namespace std;
typedef long long ll;
const int N=200000+10;
const int mod=1e9+7;
int siz[N],n,x,y,ans[N];
vector<int> g[N];
int qpow(int a,int n)
{
	int res=1;
	while(n)
	{
		if(n&1)res=1ll*res*a%mod;
		a=1ll*a*a%mod;
		n>>=1;
	}
	return res;
}
int inv(int x){
	return qpow(x,mod-2);
}
int dfs(int u,int p)
{
	siz[u]=1;
	for(int v:g[u]){
		if(v==p)continue;
		siz[u]+=dfs(v,u);
	}
	return siz[u];
}
void dfs1(int u,int p)
{
	for(int v:g[u]){
		if(v==p)continue;
		ans[v]=1ll*ans[u]*inv(siz[v])%mod*(n-siz[v])%mod;
		dfs1(v,u);
	}
}
int main()
{
    //freopen("a.in","r",stdin);
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n;
    for(int i=1;i<n;i++){
    	cin>>x>>y;
    	g[x].push_back(y);
    	g[y].push_back(x);
    }
    dfs(1,0);
    ans[1]=1;
    int res=1;
    for(int i=1;i<=n;i++)res=1ll*res*i%mod;
    for(int i=1;i<=n;i++)ans[1]=1ll*ans[1]*siz[i]%mod;
    dfs1(1,0);
	for(int i=1;i<=n;i++){
		cout<<1ll*res*inv(ans[i])%mod<<"\n";
	}
    return 0;
}