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;
   #define int long long
const int MAX = 510000;
const int MOD = 1000000007;
//
int fac[MAX], finv[MAX], inv[MAX],n;
int ans[200000],ans2[200000];vector<int> hen[200000];bool used[200000];
// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
int COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
int dfs(int a){
     int t=0,k,p=1;used[a]=true;
     for(int i=0;i<hen[a].size();i++)if(!used[hen[a][i]]){
         p=(p* dfs(hen[a][i]))%MOD;
          t+=ans[hen[a][i]];
          p=(p*COM(t,ans[hen[a][i]]))%MOD;
     }
     ans[a]=t+1;
     return p;

}
void dfs2(int a){
     used[a]=true;
     for(int i=0;i<hen[a].size();i++)if(!used[hen[a][i]]){
          ans2[hen[a][i]]=(ans2[a]*ans[hen[a][i]]%MOD*inv[n-ans[hen[a][i]]])%MOD;
          dfs2(hen[a][i]);
     }
     return;
}
     signed main(){
          COMinit();
  
       cin>>n;
     int a,b;
     for(int i=0;i<n-1;i++){
          cin>>a>>b;a--;b--;
          hen[a].push_back(b);
          hen[b].push_back(a);
     }
     fill(used,used+n,0);
     ans2[0]=dfs(0);
     fill(used,used+n,0);
     dfs2(0);
     for(int i=0;i<n;i++)cout<<ans2[i]<<endl;


   }
         


        
