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<algorithm>
#include<cstring>
#define ll long long
using namespace std;
const ll MOD=1e9+7;
struct Edge
{
	int to;
	int nxt;
}e[400005];
int n,m,edgenum,head[200005],size[200005],pa[200005];
ll fact[200005],inv[200005],f[200005],g[200005];
void add(int u,int v)
{
	e[++edgenum].to=v;
	e[edgenum].nxt=head[u];
	head[u]=edgenum;
}
ll quick_pow(ll x,ll a)
{
	ll ans=1;
	while(a)
	{
		if(a&1)ans=ans*x%MOD;
		x=x*x%MOD;
		a>>=1;
	}
	return ans;
}
void dfs(int node)
{
	size[node]=1;
	f[node]=1;
	for(int hd=head[node];hd;hd=e[hd].nxt)
	{
		int to=e[hd].to;
		if(to==pa[node])continue;
		pa[to]=node;
		dfs(to);
		size[node]+=size[to];
		f[node]=f[node]*f[to]%MOD*inv[size[to]]%MOD;
	}
}
void dfs2(int node)
{
	if(node!=1)g[node]=g[pa[node]]*f[pa[node]]%MOD*quick_pow(f[node],MOD-2)%MOD*size[node]%MOD*inv[n-size[node]]%MOD;
	else g[node]=1;
	for(int hd=head[node];hd;hd=e[hd].nxt)
	{
		int to=e[hd].to;
		if(to==pa[node])continue;
		pa[to]=node;
		dfs2(to);
	}
}
int main()
{
	scanf("%d",&n);
	fact[0]=1;
	for(int i=1;i<=n;i++)fact[i]=fact[i-1]*i%MOD;
	for(int i=1;i<=n;i++)inv[i]=quick_pow(i,MOD-2);
	inv[0]=1;
	for(int i=1;i<n;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		add(u,v);
		add(v,u);
	}
	dfs(1);
	dfs2(1);
	for(int i=1;i<=n;i++)printf("%lld\n",f[i]*g[i]%MOD*fact[n-1]%MOD);
	return 0;
}