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>
using namespace std;
const int N=2e5+9;
const int mod=1e9+7;
#define ll long long 
int dp[N],ans[N];
vector<int>g[N];
int siz[N];
int n;
int frac[N],inv[N];
void init(){
    frac[0]=frac[1]=inv[0]=inv[1]=1;
    for(int i=2;i<N;i++)frac[i]=1ll*frac[i-1]*i%mod,inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
    for(int i=2;i<N;i++)inv[i]=1ll*inv[i]*inv[i-1]%mod;
}
int C(int a,int b){return 1ll*frac[a]*inv[b]%mod*inv[a-b]%mod;}
void dfs1(int u,int fa){
   dp[u]=1;
   for(auto v:g[u]){
      if(v==fa)continue;
      dfs1(v,u);
      siz[u]+=siz[v];
      dp[u]=1ll*dp[v]*C(siz[u],siz[v])%mod*dp[u]%mod;
   }
   siz[u]++;
}
int ny(int a){
   if(a==1)return 1;
   else return 1ll*(mod-mod/a)*ny(mod%a)%mod;
}
void dfs2(int u,int fa){
   for(auto v:g[u]){
      if(v==fa)continue;
      dp[v]=1ll*dp[v]*C(n-1,n-siz[v])%mod*dp[u]%mod*ny(1ll*C(n-1,siz[v])*dp[v]%mod)%mod;
      dfs2(v,u);
   }
}
int main(){
   ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
   init();
   cin>>n;
   for(int i=1,a,b;i<n;i++)cin>>a>>b,g[a].push_back(b),g[b].push_back(a);
   dfs1(1,0);dfs2(1,0);
   for(int i=1;i<=n;i++)cout<<dp[i]<<endl;
   return 0;
}