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 LL long long
using namespace std;

const int maxn=1e6+7,mod=1e9+7;
LL fac[maxn],inv[maxn],f[maxn];
int n,siz[maxn];
vector<int> to[maxn];

LL poww(LL a,LL b)
{
	LL ans=1;
	while(b) {
		if(b&1) ans=(ans*a)%mod;
		a=a*a%mod;
		b>>=1;
	}
	return ans;
}

void dfs(int x,int fa)
{
	siz[x]=1;
	for(int i=0;i<to[x].size();++i) {
		int t=to[x][i];
		if(t==fa) continue;
		dfs(t,x);
		siz[x]+=siz[t];
	}
}

void dfs2(int x,int fa)
{
	for(int i=0;i<to[x].size();++i) {
		int t=to[x][i];
		if(t==fa) continue;
		f[t]=(f[x]*siz[t]%mod)*inv[n-siz[t]]%mod;
		dfs2(t,x);
	}
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;++i) {
		int a,b;
		scanf("%d %d",&a,&b);
		to[a].push_back(b);
		to[b].push_back(a);
	}
	fac[0]=inv[0]=inv[1]=1;
	for(int i=1;i<maxn;++i) fac[i]=fac[i-1]*i%mod;
	for(int i=2;i<maxn;i++) inv[i]=inv[mod%i]*(mod-mod/i)%mod;
	dfs(1,0);
	f[1]=fac[n];
	for(int i=1;i<=n;++i) 
		f[1]=(f[1]*inv[siz[i]])%mod;
	dfs2(1,0);
	for(int i=1;i<=n;++i) printf("%lld\n",f[i]);
}