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<cstdio>
#include <stdio.h>
#include<algorithm>
#include<cstring>
#include <string>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<map>
#include<vector>
#include<bits/stdc++.h>
#include <set>
#define ll   long long
#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#define inf 0x3f3f3f3f3f3f3f3f
#define pi 3.1415926535898

#define N 200010
using namespace std;
vector <int >g[N];
const ll mod=1e9+7;
int n,x,y;
ll power(ll a,ll b)
{
    return b?power(a*a%mod,b/2)*(b%2?a:1)%mod:1;
}
ll p[N],f[N],c[N];
void dfs(int u)
{
    c[u]=1;
    f[u]=1;
    for(auto v: g[u])
    {
        if(!c[v])
        {
            dfs(v);
            c[u]+=c[v];
            p[v]=u;
            f[u]=f[u]*f[v]%mod;
        }
    }
    f[u]=f[u]*c[u]%mod;
}
void calculate(int u)
{
    for(auto v:g[u])
    {
        if(v!=p[u])
        {
            f[v]=f[u]*(n-c[v])%mod*power(c[v],mod-2)%mod;
            calculate(v);
        }
    }
}
int main()
{
    cin>>n;
    for(int i=0; i<n-1; i++)
    {
        cin>>x>>y;
        g[x].push_back(y);
        g[y].push_back(x);
    }
    ll S=1;
    for(int i=1; i<=n; i++)
        S=S*i%mod;
    dfs(1);
    calculate(1);
    for(int i=1; i<=n; i++)
        cout<<S*power(f[i],mod-2)%mod<<endl;
    return 0;
}
