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 int long long
#define f first
#define s second
using namespace std;
const int N=2e5+9,mod=1e9+7;
int n,sz[N],fac[N],ifac[N],f[N],h[N];
vector <int> g[N];
int pw(int a,int t=mod-2){
    int x=1;
    while(t>0){
        if(t&1)x=x*a%mod;
        a=a*a%mod,t/=2;
    }
    return x;
}
void dfs(int v,int p){
    f[v]=1;
    for(auto u:g[v]){
        if(u==p)continue;
        dfs(u,v);
        f[v]=f[v]*f[u]%mod;
        f[v]=f[v]*ifac[sz[u]]%mod;
        sz[v]+=sz[u];
    }
    f[v]=f[v]*fac[sz[v]]%mod;
    sz[v]++;
}
void dfs2(int v,int p){
    h[v]=f[p]*pw(f[v])%mod*fac[sz[v]]%mod;
    h[v]=h[v]*ifac[sz[p]-1]%mod*fac[n-sz[v]-1]%mod;
    h[v]=h[v]*h[p]%mod*ifac[n-sz[p]]%mod;
    if(v==1)h[v]=1;
    for(auto u:g[v]){
        if(u==p)continue;
        dfs2(u,v);
    }
}
int32_t main(){
    if(fopen("input.txt", "r"))
    freopen("input.txt", "r", stdin),
    freopen("output.txt", "w", stdout);
    ios::sync_with_stdio(0),cin.tie(0);
    fac[0]=ifac[0]=1;
    for(int i=1;i<N;i++){
        fac[i]=i*fac[i-1]%mod;
        ifac[i]=pw(fac[i]);
    }
    cin >> n;
    for(int i=1;i<n;i++){
        int u,v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1,0);
    sz[0]=sz[1];
    f[0]=f[1];
    h[0]=1;
    dfs2(1,0);
    for(int i=1;i<=n;i++){
        int x=fac[n-1]*f[i]%mod*h[i]%mod;
        x=x*ifac[sz[i]-1]%mod*ifac[n-sz[i]]%mod;
        cout << x << '\n';
    }
}