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 ll long long
using namespace std;
const int N=2e5+5,mod=1e9+7;
vector<int>v[N];
ll res=1,f[N],inv[N],invf[N];
int n,si[N],ans[N];
void dfs(int x,int pre)
{
    si[x]=1;
    for(int i=0;i<v[x].size();i++)
    {
        if(v[x][i]==pre) continue;
        dfs(v[x][i],x);
        si[x]+=si[v[x][i]];
        res=res*invf[si[v[x][i]]]%mod;
    }
    res=res*f[si[x]-1]%mod;
}
void solve(int x,int pre,ll res)
{
    ans[x]=res;
    for(int i=0;i<v[x].size();i++)
    {
        if(v[x][i]==pre) continue;
        solve(v[x][i],x,res*inv[n-si[v[x][i]]]%mod*si[v[x][i]]%mod);
    }
}
int main()
{
    scanf("%d",&n);
    f[0]=f[1]=invf[0]=invf[1]=1,inv[1]=1;
    for(int i=2;i<N;i++)
    {
        f[i]=f[i-1]*i%mod;
        inv[i]=(mod-mod/i)*inv[mod%i]%mod;
        invf[i]=invf[i-1]*inv[i]%mod;
    }
    for(int i=1;i<n;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        v[x].push_back(y);
        v[y].push_back(x);
    }
    dfs(1,0);
    solve(1,0,res);
    for(int i=1;i<=n;i++) printf("%d\n",ans[i]);
}
