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>
using namespace std;
const int MAXN = 2e5 + 10;
struct EdgeNode{
	int next,to;
}edge[MAXN<<1];
int head[MAXN],cnt=1,N;
void add(int u,int v){
	cnt++; edge[cnt].next=head[u]; head[u] = cnt; edge[cnt].to = v;
}
void insert(int u,int v){
	add(u,v); add(v,u); 
}
const int MOD = 1e9 + 7;
#define LL long long
int sz[MAXN];
LL fac[MAXN],ans[MAXN],sum;

LL qpow(LL x, LL y){
	LL re = 1;
	for (; y; y>>=1,x=x*x%MOD)
		if (y&1) re = re*x%MOD;
	return re;
}
void dfs(int u,int last){
	sz[u] = 1;
	fac[u] = 1;
	for (int i=head[u]; i; i=edge[i].next){
		int v = edge[i].to;
		if (v == last) continue;
		dfs(v,u);
		sz[u] += sz[v];
		fac[u] = fac[u]*fac[v]%MOD;
	}
	fac[u] = fac[u]*sz[u]%MOD;
}

void dfs2(int u,int last,LL p){
	LL t = p;
	for (int i=head[u]; i; i=edge[i].next){
		int v = edge[i].to;
		if (v == last) continue;
		t = t*fac[v]%MOD;
	}
	ans[u] = sum*qpow(t,MOD-2)%MOD*qpow(N,MOD-2)%MOD;
	for (int i=head[u]; i; i=edge[i].next){
		int v = edge[i].to;
		if (v == last) continue;
		p = p*fac[v]%MOD;
	}
	for (int i=head[u]; i; i=edge[i].next){
		int v = edge[i].to;
		if (v == last) continue;
		dfs2(v,u,p*qpow(fac[v],MOD-2)%MOD*(N-sz[v])%MOD);
	}
}

int main(){
	scanf("%d",&N);
	for (int i=1; i<N; i++){
		int u,v; scanf("%d%d",&u,&v);
		insert(u,v);
	}	
	sum = 1LL;
	for (int i=1; i<=N; i++) sum = sum*i%MOD;
	dfs(1,0);
	dfs2(1,0,1);
//	for (int i=1; i<=N; i++) printf("%d ",fac[i]); puts("");
	for (int i=1; i<=N; i++) printf("%I64d\n",ans[i]);
	return 0;
}