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>
typedef long long ll;
const int mod=1e9+7;
const int N=2e5+5;
ll po(ll a,ll b)
{
	ll ans=1;
	while(b)
	{
		if(b&1) ans=ans*a%mod;
		a=a*a%mod,b>>=1;
	}
	return ans;
}
ll inv(ll x){return po(x,mod-2);}
ll jc[N<<1],jcinv[N<<1];
inline ll c(ll a,ll b)
{return jc[a]*jcinv[b]%mod*jcinv[a-b]%mod;}



int n;
struct Edge{
	int nxt,to;
}e[N<<1];
int cnt=1,head[N];
void inline add(int u,int v)
{
	e[cnt]={head[u],v};
	head[u]=cnt++;
}

ll sz[N];//子树大小
ll f[N];//走完子树的方法数
ll ans[N];
void dfs1(int u,int fa)
{
	f[u]=1;
	for(int i=head[u];i;i=e[i].nxt)
	{
		int v=e[i].to;
		if(v==fa) continue;
		dfs1(v,u);
		sz[u]+=sz[v];
		f[u]=f[u]*c(sz[u],sz[v])%mod*f[v]%mod;
	}
	sz[u]++;
}
void dfs2(int u,int fa,ll ffa)
{
	ans[u]=f[u]*c(n-1,n-sz[u])%mod*ffa%mod;
	for(int i=head[u];i;i=e[i].nxt)
	{
		int v=e[i].to;
		if(v==fa) continue;
		dfs2(v,u,ans[u]*inv(f[v])%mod*inv(c(n-1,sz[v]))%mod);
	}
}

int main()
{
	jc[0]=jcinv[0]=jc[1]=jcinv[1]=1;
	for(int i=2;i<(N<<1);i++)
	{
		jc[i]=jc[i-1]*i%mod;
		jcinv[i]=po(jc[i],mod-2);
	}
	scanf("%d",&n);
	for(int i=1,u,v;i<n;i++)
	{
		scanf("%d%d",&u,&v);
		add(u,v);
		add(v,u);
	}
	dfs1(1,0);
	dfs2(1,0,1);
	for(int i=1;i<=n;i++)
	{
		printf("%lld\n",ans[i]);
	}
	return 0;
}
