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 <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;
const int mo=1e9+7;
const int maxn=1e6+10;
int n,st[maxn],nt[maxn],to[maxn],si[maxn],topt;
long long pp[maxn],inv[maxn],dp[maxn],ans[maxn];
inline void add(int x,int y){to[++topt]=y; nt[topt]=st[x]; st[x]=topt;}
long long po(long long a,long long b)
{
	if (!b) return 1;
	if (b==1) return a;
	long long c=po(a,b/2);
	if (b&1) return c*c%mo*a%mo;else return c*c%mo;
}
void dfs1(int x,int fa)
{
	dp[x]=1; si[x]=1; int p=st[x];
	while (p)
	{
		if (to[p]!=fa) {dfs1(to[p],x); si[x]+=si[to[p]]; dp[x]=dp[x]*dp[to[p]]%mo*inv[si[to[p]]]%mo;}
		p=nt[p];
	}
	dp[x]=dp[x]*pp[si[x]-1]%mo;
}
void dfs2(int x,int fa,long long now)
{
	ans[x]=dp[x]*inv[si[x]-1]%mo*now%mo*inv[n-si[x]]%mo*pp[n-1]%mo;
	int p=st[x];
	while (p)
	{
		if (to[p]!=fa) 
		{
			dfs2(to[p],x,ans[x]*inv[n-1]%mo*po(dp[to[p]],mo-2)%mo*pp[si[to[p]]]%mo*pp[n-si[to[p]]-1]%mo);
		}
		p=nt[p];
	}
}
int main()
{
	scanf("%d",&n);
	for (int i=1;i<n;i++) {int x,y; scanf("%d%d",&x,&y); add(x,y); add(y,x);}
	pp[0]=1; for (int i=1;i<=n;i++) pp[i]=pp[i-1]*i%mo;
	inv[n]=po(pp[n],mo-2); for (int i=n-1;i>=0;i--) inv[i]=inv[i+1]*(i+1)%mo;
	dfs1(1,0); dfs2(1,0,1);
	for (int i=1;i<=n;i++) printf("%lld\n",ans[i]);
return 0;
}