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+5,mod=1e9+7;
#define ll long long
ll f[N];
ll inv[N];
vector<int> adj[N];
pair<ll,ll> dp[N];
ll ans[N];
ll fp(ll b,ll p)
{
    if(p==0) return 1;
    ll t=fp(b,p/2);
    if(p%2) return t*t%mod*b%mod;
    return t*t%mod;
}
ll ncr(int n,int r)
{
    return f[n]*inv[r]%mod*inv[n-r]%mod;
}
void Merge(int u,int v)
{
    dp[v].first++;
    dp[u].second*=dp[v].second*ncr(dp[u].first+dp[v].first,dp[u].first)%mod;
    dp[u].second%=mod;
    dp[u].first+=dp[v].first%mod;
    dp[u].first%=mod;
    dp[v].first--;
}
void DeMerge(int u,int v)
{
    dp[v].first++;
    dp[u].first-=dp[v].first%mod;;
    dp[u].first=(dp[u].first+mod)%mod;
    dp[u].second=dp[u].second*fp(dp[v].second*ncr(dp[u].first+dp[v].first,dp[u].first)%mod,mod-2)%mod;
    dp[v].first--;
}
void solve(int u,int p)
{
    dp[u].first=0;
    dp[u].second=1;
    for(auto v:adj[u])
        if(v!=p)
        {
            solve(v,u);
            Merge(u,v);
        }
}
void dfs(int u,int p)
{
    ans[u]=dp[u].second;
    for(auto v:adj[u])
        if(v!=p)
        {
            DeMerge(u,v);
            Merge(v,u);
            dfs(v,u);
            DeMerge(v,u);
            Merge(u,v);
        }
}
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    f[0]=1;
    for(int i=1;i<N;i++) f[i]=(f[i-1]*i)%mod;
    for(int i=0;i<N;i++) inv[i]=fp(f[i],mod-2);
    int n;
    cin>>n;
    for(int i=1;i<n;i++)
    {
        int u,v;
        cin>>u>>v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    solve(1,-1);
    dfs(1,-1);
    for(int i=1;i<=n;i++)
        cout<<ans[i]<<'\n';
    return 0;
}
