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;
const int MD=1e9+7;
vector<int> G[N];
int f[N],ans[N];
int n,a,b;
int quick_pow(int x,int y) {
    int ans=1;
    while(y) {
        if(y&1) ans=1LL*ans*x%MD;
        y>>=1;
        x=1LL*x*x%MD;
    }
    return ans;
}
void dfs(int u,int fa) {
    for(auto v:G[u]) {
        if(v==fa) continue;
        dfs(v,u);
        f[u]+=f[v];
    }
    f[u]++;
}
void cal(int u,int fa) {
    for(auto v:G[u]) {
        if(v==fa) continue;
        ans[v]=1LL*ans[u]*quick_pow(f[v],MD-2)%MD*(n-f[v])%MD;
        cal(v,u);
    }
}
int main() {
    //freopen("in.txt","r",stdin);
    scanf("%d",&n);
    for(int i=1;i<n;i++) {
        scanf("%d%d",&a,&b);
        G[a].push_back(b);
        G[b].push_back(a);
    }
    dfs(1,-1);
    ans[1]=1;
    for(int i=1;i<=n;i++) {
        ans[1]=1LL*ans[1]*f[i]%MD;
    }
    cal(1,-1);
    int t=1;
    for(int i=1;i<=n;i++) t=1LL*t*i%MD;
    for(int i=1;i<=n;i++) {
        printf("%d\n",1LL*t*quick_pow(ans[i],MD-2)%MD);
    }
    return 0;
}