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;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<int,ll> pil;
typedef pair<ll,int> pli;
typedef vector<int> vi;
#define rep(i,n) for (int i=0;i<n;++i)
#define REP(i,n) for (int i=1;i<=n;++i)
#define all(x) x.begin(),x.end()
#define mp make_pair
#define pb push_back
#define F first
#define S second
#define read(x) scanf("%d",&x)

const ll P=1e9+7;
const int MAXN=200005;
inline ll inv(ll x){
	ll exp=P-2,a=1,tmp=x;
	while(exp){
		if (exp&1) a=a*tmp%P;
		tmp=tmp*tmp%P;
		exp>>=1;
	}
	return a;
}

int n;
ll fact[MAXN];
vector<int> G[MAXN];
int fa[MAXN],sz[MAXN];
ll dp[MAXN];
inline void dfs(int u,int pa){
	dp[u]=sz[u]=1;
	for (int i=0;i<G[u].size();++i){
		int v=G[u][i];
		if (v!=pa){
			dfs(v,u);sz[u]+=sz[v];dp[u]=dp[u]*dp[v]%P*inv(fact[sz[v]])%P;
		}
	}
	dp[u]=dp[u]*fact[sz[u]-1]%P;
	return;
}
inline void reroot(int u,int pa){
	for (int i=0;i<G[u].size();++i){
		int v=G[u][i];
		if (v!=pa){
			dp[v]=dp[u]*sz[v]%P*inv(n-sz[v])%P;reroot(v,u);
		}
	}
}

int main(){
	ios::sync_with_stdio(false);
	cin>>n;
	fact[0]=1;
	REP(i,n) fact[i]=fact[i-1]*i%P;
	REP(i,n-1){
		int u,v;cin>>u>>v;
		G[u].pb(v);G[v].pb(u);
	}
	dfs(1,0);
	reroot(1,0);
	REP(i,n) cout<<dp[i]%P<<endl;
	return 0; 
}