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 maxn 1000005
using namespace std;
const int mod = 1000000007;
int n;
vector<int> g[maxn];
int fac[maxn],inv[maxn];
int dp[maxn],sz[maxn],Ans[maxn];
int fastpow(int a,int p)
{
	int ans=1;
	for(;p;p>>=1,a=1ll*a*a%mod)if(p&1)ans=1ll*ans*a%mod;
	return ans; 
}
int C(int x,int y)
{
	return 1ll*fac[x]*inv[y]%mod*inv[x-y]%mod;
}
void dfs(int u,int f)
{
	dp[u]=1;
	for(int v:g[u])if(v!=f)
	{
		dfs(v,u);
		dp[u]=1ll*dp[v]*dp[u]%mod*C(sz[u]+sz[v],sz[v])%mod;
		sz[u]+=sz[v];
	}
	sz[u]+=1;
}
void dfs2(int u,int f)
{
	for(int v:g[u])if(v!=f)
	{
		int t=1ll*Ans[u]*fastpow(dp[v],mod-2)%mod*fastpow(C(n-1,sz[v]),mod-2)%mod;
		Ans[v]=1ll*Ans[u]*fastpow(C(n-1,sz[v]),mod-2)%mod*C(n-1,sz[v]-1)%mod; 
		dfs2(v,u);
	}
}
int main()
{
	fac[0]=inv[0]=1;
	for(int i=1;i<=1000000;++i)fac[i]=1ll*fac[i-1]*i%mod,inv[i]=fastpow(fac[i],mod-2);
	scanf("%d",&n);
	for(int i=1;i<n;++i)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		g[u].push_back(v);g[v].push_back(u);
	}
	dfs(1,0);
	Ans[1]=dp[1];
	dfs2(1,0);
	for(int i=1;i<=n;++i)printf("%d\n",Ans[i]); 
}