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;
const long long MAXN = 3e5+5;
const long long MOD = 1e9+7;
long long n;
long long sz[MAXN];
long long fact[MAXN];
long long dp[MAXN];
long long dp2[MAXN];
long long invfact[MAXN];
long long inv[MAXN];
vector<long long> v1[MAXN];
void dfs(long long curr,long long par){
    sz[curr] = 1;
    dp[curr] = 1;
    
    for(long long x:v1[curr]){
        if(x!=par){
            dfs(x,curr);
            sz[curr]+=sz[x];
            //cout<<dp[x]<<endl;
            dp[curr]*=((invfact[sz[x]]%MOD)*(dp[x]))%MOD;
            dp[curr]%=MOD;
        }
    }
 //   cout<<curr<<" "<<fact[sz[curr]-1]<<" "<<dp[curr]<<endl;
    dp[curr]*=fact[sz[curr]-1];
    dp[curr]%=MOD;
    
}
void dfs2(long long curr,long long par){
    if(curr!=1){
        dp[curr] = (((dp[par]*sz[curr])%MOD)*inv[n-sz[curr]])%MOD;
    }
    for(long long x:v1[curr]){
        if(x!=par){
            dfs2(x,curr);
        }
    }
}
int main() {
    cin>>n;
    for(long long i=1;i<n;i++){
        long long x,y;
        cin>>x>>y;
        v1[x].push_back(y);
        v1[y].push_back(x);
        
    }
    inv[1] = 1;
    invfact[1] = 1;
    inv[0]= 1;
    invfact[0] = 1;
    fact[0] = 1;
    fact[1] = 1;
    for(long long i=2;i<=200000;i++){
        fact[i] = 1LL*fact[i-1]*i;
        fact[i]%=MOD;
        
        inv[i]=1LL*(MOD-(MOD/i))*inv[MOD%i]%MOD;
        invfact[i] = invfact[i-1]*inv[i]%MOD;
        //cout<<fact[i]<<" "<<inv[i]<<" "<<invfact[i]<<endl;
    }
    dfs(1,1);
    dfs2(1,1);
    for(long long i=1;i<=n;i++){
        cout<<dp[i]<<endl;
    }
}