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 maxn=2e5;
typedef long long ll;
const ll mod=1e9+7;
ll fac[maxn+5],inv[maxn+5];
vector<int> g[maxn+5];
ll dp[maxn+5],ans[maxn+5];
int sz[maxn+5];
int n;
ll C(int n,int m)
{
    if(n<m) return 0;
    return fac[n]*inv[m]%mod*inv[n-m]%mod;
}
void cal(ll &f,int n,ll g,int m)
{
    f=C(n+m,n)*f%mod*g%mod;
}
void mul(ll &a,ll b)
{
    a=a*b%mod;
}
void dfs(int k,int fa)
{
    sz[k]=1;
    dp[k]=1;
    for(auto u:g[k])
        if(u!=fa)
        {
            dfs(u,k);
            mul(dp[k],dp[u]*inv[sz[u]]%mod);
            sz[k]+=sz[u];
        }
    // mul(dp[k],inv[sz[k]]);
}
ll Pow(ll a,ll b,ll mod)
{
    ll ans=1;
    while(b)
    {
        if(b&1) ans=ans*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return ans;
}
ll Inv(ll x)
{
    return Pow(x,mod-2,mod);
}
void dfs1(int k,int fa,ll num)
{
    // if(k==4) cout<<"ce "<<num<<" "<<sz[k]<<endl;
    // if(k==2) cout<<"ce "<<num<<endl;
    ans[k]=dp[k]*num%mod*fac[n-1]%mod;
    if(ans[k]<0) ans[k]+=mod;
    for(auto u:g[k])
        if(u!=fa)
            dfs1(u,k,dp[k]*Inv(dp[u])%mod*num%mod*inv[n-sz[u]]%mod*sz[u]%mod);
}
int main()
{

    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n;
    for(int i=1;i<n;++i)
    {
        int u,v;
        cin>>u>>v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    fac[0]=1;
    for(int i=1;i<=n;++i) fac[i]=fac[i-1]*i%mod;
    inv[0]=inv[1]=1;
    for(int i=2;i<=n;++i) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
    // for(int i=1;i<=n;++i) inv[i]=inv[i-1]*inv[i]%mod;
    dfs(1,0);

    dfs1(1,0,1);
    for(int i=1;i<=n;++i) printf("%lld\n",ans[i]);
    return 0;
}