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 int long long
#define MAXN 200005
#define mod 1000000007
using std::vector;
int n,i,x,f[MAXN],size[MAXN];
vector<int> G[MAXN];
bool vis[MAXN];
inline void read(int &x)
{
	short negative=1;
    x=0;
    char c=getchar();
    while(c<'0' || c>'9')
    {
    	if(c=='-')
			negative=-1;
		c=getchar();
	}
    while(c>='0' && c<='9')
        x=(x<<3)+(x<<1)+(c^48),c=getchar();
    x*=negative;
}
inline void print(int x)
{
    if (x<0)
        putchar('-'),x=-x;
    if (x>9)
        print(x/10);
    putchar(x%10+'0');
}
inline int ksm(int a,int n,int p)
{
	int Ret(1);
	while (n)
	{
		if (n&1)
			(Ret*=a)%=p;
		(a*=a)%=p,n>>=1;
	}
	return Ret;
}
inline void dfs(int u)
{
	vis[u]=true,size[u]=1;
	for (int i=0;i<G[u].size();i++)
	{
		int v=G[u][i];
		if (vis[v])
			continue;
		dfs(v);
		size[u]+=size[v];
	} 
	(f[1]*=size[u])%=mod;
}
inline void dfs2(int u)
{
	vis[u]=true;
	for (int i=0;i<G[u].size();i++)
	{
		int v=G[u][i];
		if (vis[v])
			continue;
		f[v]=f[u]*((n-size[v])*ksm(size[v],mod-2,mod)%mod)%mod;
		dfs2(v);
	}
}
signed main(void)
{
	read(n);
	for (i=1;i<n;i++)
	{
		int u,v;
		read(u),read(v);
		G[u].push_back(v);
		G[v].push_back(u);
	}
	f[1]=1;
	dfs(1);
	memset(vis,false,sizeof(vis));
	dfs2(1);
	for (i=x=1;i<=n;i++)
		(x*=i)%=mod;
	for (i=1;i<=n;i++)
		print((x*ksm(f[i],mod-2,mod))%mod),putchar('\n');
	return 0;
}