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 <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <queue>
#define xx first
#define yy second
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N=3e5+10;
const int M=1e6+10;
const int inf=0x3f3f3f3f;
const LL INF=0x3f3f3f3f3f3f3f3f;
const int mod=1e9+7;
int n,siz[N];
vector<int> g[N];
LL fac[N],ans[N],mul[N];

LL qpow(LL x,LL k){
	LL res=1;
	while(k){
		if(k&1) res=res*x%mod;
		k>>=1;
		x=x*x%mod;
	}
	return res;
}

void predfs(int u,int fa){
	siz[u]=mul[u]=1;
	for(int v:g[u]){
		if(v==fa) continue;
		predfs(v,u);
		siz[u]+=siz[v];
		mul[u]=mul[u]*mul[v]%mod;
	}
	mul[u]=mul[u]*siz[u]%mod;
}

void dfs(int u,int fa,LL premul){
	LL down=premul;
	for(int v:g[u]) if(v!=fa) down=down*mul[v]%mod;
	ans[u]=fac[n-1]*qpow(down,mod-2)%mod;
	for(int v:g[u]) if(v!=fa) premul=premul*mul[v]%mod;
	for(int v:g[u]) if(v!=fa) dfs(v,u,premul*qpow(mul[v],mod-2)%mod*(n-siz[v])%mod);
}

int main(){
	scanf("%d",&n);
	for(int i=1,u,v;i<n;i++){
		scanf("%d%d",&u,&v);
		g[u].push_back(v);
		g[v].push_back(u);
	}
	fac[0]=1;
	for(int i=1;i<=n;i++) fac[i]=fac[i-1]*i%mod;
	predfs(1,0);
	dfs(1,0,1);
	for(int i=1;i<=n;i++) printf("%lld\n",ans[i]);
	return 0;
}