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 <algorithm>
#include <vector>
#include <cstring>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cstdio>
#include <cmath>
#include <limits.h>
using namespace std;
typedef long long LL;
const int N=2e5+5;
const LL M=1e9+7;

int n,u,v,sz[N];
LL cnt[N],ans[N];
vector<int> G[N];
LL p[N],inv[N];

LL exgcd(LL a,LL b,LL &x,LL &y) {
	LL d=a;
	if(b!=0) {
		d=exgcd(b,a%b,y,x);
		y-=(a/b)*x;
	}
	else {
		x=1;y=0;
	}
	return d;
}

LL mod_inv(LL a,LL m) {
	LL x,y;
	exgcd(a,m,x,y);
	return (m+x%m)%m;
}

LL C(int n, int m) {
	return p[n]*inv[n-m]%M*inv[m]%M;
}

void init() {
	p[0]=1;
	inv[0]=1;
	for(int i=1;i<=n;i++) {
		p[i]=p[i-1]*i%M;
		inv[i]=mod_inv(p[i],M);
	}
}

void dfs(int u,int f) {
	sz[u]=1;
	cnt[u]=1;
	for(int v:G[u]) {
		if(v==f) continue;
		dfs(v,u);
		sz[u]+=sz[v];
		cnt[u]=cnt[u]*cnt[v]%M*inv[sz[v]]%M;
	}
	cnt[u]=cnt[u]*p[sz[u]-1]%M;
}

void dfs2(int u,int f) {
	if(u==1) {
		ans[u]=cnt[u];
	}
	else {
		//cout<<"gou "<<u<<endl;
		//printf("ans[%d]=%lld\n", f,ans[f]);
		ans[u]=ans[f]*mod_inv(C(n-1,sz[u])*cnt[u]%M,M)%M*inv[n-sz[u]]%M;
		//printf("aaa %lld %lld %lld %lld %lld\n",ans[f],C(n-1,sz[u]),cnt[u],mod_inv(C(n-1,sz[u])*cnt[u]%M,M),inv[n-sz[u]]);
		for(int v:G[u]) {
			if(v==f) continue;
			ans[u]=ans[u]*cnt[v]%M*inv[sz[v]]%M;		
		}
		ans[u]=ans[u]*p[n-1]%M;
	}
	for(int v:G[u]) {
		if(v==f) continue;
		dfs2(v,u);
	}
}

int main() {
	scanf("%d",&n);
	init();
	for(int i=1;i<n;i++) {
		scanf("%d %d",&u,&v);
		G[u].push_back(v);
		G[v].push_back(u);
	}
	dfs(1,0);
	dfs2(1,0);
	for(int i=1;i<=n;i++) {
		printf("%lld\n",ans[i]);
	}
}
