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<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,m,MOD=1e9+7,a,b,ans,bik[200001],dp[200000],par[200000],num[200000],dp2[200000];
vector<ll> v[200000];
ll f(ll x,ll y){
    if(y==0)return 1;
    if(y%2==0)return f(x,y/2)*f(x,y/2)%MOD;
    return x*f(x,y-1)%MOD;
}
ll comb(ll x, ll y){
    return bik[x]*f(bik[x-y],MOD-2)%MOD*f(bik[y],MOD-2)%MOD;
}
void dfs(int a){
    for(int x:v[a]){
        if(x==0||par[x]!=-1)continue;
        par[x]=a;
        dfs(x);
        num[a]+=num[x];
        //cout<<comb(num[a],num[x])<<" "<<dp[a]<<" <<endl;
        dp[a]*=dp[x]*comb(num[a],num[x])%MOD;
        dp[a]%=MOD;
    }
    num[a]++;
    return ;
}
void dfs2(int a){
    if(a==0)dp2[a]=dp[a];
    else{
        dp2[a]=dp[a];
        dp2[a]*=comb(n-1,num[a]-1)*dp2[par[a]]%MOD*f(dp[a]*comb(n-1,num[a])%MOD,MOD-2)%MOD;
        //cout<<dp[a]<<" "<<dp2[a]<<" "<<dp2[par[a]]<<" "<<f(dp[a]*comb(num[par[a]]-1,num[a])%MOD,MOD-2)%MOD<<" "<<dp[a]*comb(num[par[a]]-1,num[a])<<endl;
        dp2[a]%=MOD;
    }
    for(int x:v[a]){
        if(x==par[a])continue;
        dfs2(x);
    }
    return ;
}
int main(void){
    cin>>n;
    bik[0]=1;
    for(ll i=1;i<=200000;i++){
        bik[i]=bik[i-1]*i;
        bik[i]%=MOD;
    }
    for(int i=0;i<n-1;i++){
        cin>>a>>b;
        a--,b--;
        v[a].push_back(b);
        v[b].push_back(a);
    }
    for(int i=0;i<n;i++)par[i]=-1;
    for(int i=0;i<n;i++)dp[i]=1;
    dfs(0);
    dfs2(0);
    for(int i=0;i<n;i++){
        cout<<dp2[i]<<endl;
    }
    
}
