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 pf printf
#define sc(x) scanf("%d",&x);
#define scs(x) scanf("%s",x);
#define scl(x) scanf("%lld",&x);
#define rep(i,s,e) for(int i=s; i<=e; ++i)
#define dep(i,s,e) for(int i=s; i>=e; --i)
typedef long long ll;
using namespace std;
const ll mod=1e9+7; 
vector<int>tree[200005];
int n;
ll ans,sz[200005],d[200005];
ll pow_mod(ll a, ll b, ll p){//a的p-2次方求余p； p为mod； 
    ll ret = 1;
    while(b){
        if(b & 1) ret = (ret * a) % p;
        a = (a * a) % p;
        b >>= 1;
    }
    return ret;
}
void cint(){
	sc(n);
	rep(i,1,n-1){
		int u,v;sc(u);sc(v);
		tree[u].push_back(v);tree[v].push_back(u);
	}d[1]=1;
	rep(i,1,n)(d[1]*=i)%=mod;
}
void dfs1(int u,int fa){
	sz[u]=1;
	for(int i=0;i<tree[u].size();i++){
		int v=tree[u][i];
		if(v==fa)continue;
		dfs1(v,u);
		sz[u]+=sz[v];	
	}(d[1]*=pow_mod(sz[u],mod-2,mod))%=mod;
}
void dfs2(int u,int fa){
	for(int i=0;i<tree[u].size();i++){
		int v=tree[u][i];
		if(v==fa)continue;
		d[v]=d[u]*sz[v]%mod*pow_mod(n-sz[v],mod-2,mod)%mod;
		dfs2(v,u);
	}
}
void anser(){
	rep(i,1,n){
		pf("%lld\n",d[i]);
	}
}
int main(){
  	cint();
  	dfs1(1,0);
  	dfs2(1,0);
  	anser();
}