#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e5+5;
const ll mod=1e9+7;
vector<int>g[N];
int n,x,y;
ll power(ll a,ll b){return b?power(a*a%mod,b/2)*(b%2?a:1)%mod:1;}
ll p[N],s[N],f[N];

void dfs(int u){
    s[u]=1;
    f[u]=1;
    for(auto v:g[u]){
        if(!s[v]){
            dfs(v);
            s[u]+=s[v];
            p[v]=u;
            f[u]=f[u]*f[v]%mod;
        }
    }
    f[u]=f[u]*s[u]%mod;
}

void calculate(int u){
    for(auto v:g[u]){
        if(v!=p[u]){
            f[v]=f[u]*(n-s[v])%mod*power(s[v],mod-2)%mod;
            calculate(v);
        }
    }
}

int main()
{
    cin>>n;
    for(int i=0;i<n-1;i++){
        cin>>x>>y;
        g[x].push_back(y);
        g[y].push_back(x);
    }
    ll S=1;
    for(int i=1;i<=n;i++) S=S*i%mod;
    dfs(1);
    calculate(1);
    for(int i=1;i<=n;i++)
        cout<<S*power(f[i],mod-2)%mod<<endl;
    return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e5+5;
const ll mod=1e9+7;
vector<int>g[N];
int n,x,y;
ll power(ll a,ll b){return b?power(a*a%mod,b/2)*(b%2?a:1)%mod:1;}
ll p[N],s[N],f[N];

void dfs(int u){
    s[u]=1;
    f[u]=1;
    for(auto v:g[u]){
        if(!s[v]){
            dfs(v);
            s[u]+=s[v];
            p[v]=u;
            f[u]=f[u]*f[v]%mod;
        }
    }
    f[u]=f[u]*s[u]%mod;
}

void calculate(int u){
    for(auto v:g[u]){
        if(v!=p[u]){
            f[v]=f[u]*(n-s[v])%mod*power(s[v],mod-2)%mod;
            calculate(v);
        }
    }
}

int main()
{
    cin>>n;
    for(int i=0;i<n-1;i++){
        cin>>x>>y;
        g[x].push_back(y);
        g[y].push_back(x);
    }
    ll S=1;
    for(int i=1;i<=n;i++) S=S*i%mod;
    dfs(1);
    calculate(1);
    for(int i=1;i<=n;i++)
        cout<<S*power(f[i],mod-2)%mod<<endl;
    return 0;
}

```