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>
#include<unordered_set>
#include<unordered_map>
using namespace std;
#define REP(i,m,n) for(int i=(m);i<(n);i++)
#define rep(i,n) REP(i,0,n)
#define pb push_back
#define all(a) a.begin(),a.end()
#define rall(c) (c).rbegin(),(c).rend()
#define mp make_pair
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef long double ld;
const ll inf=1e9+7;
const ll mod=1e9+7;
const ll MAX=200005;
vector<ll>G[MAX];
vector<ll>ans(MAX);
vector<ll> siz(MAX);
ll rui(ll a,ll b){
    ll res=1;
    while(b>0){
        if(b&1)res=res*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return res;
}
void preDFS(ll u,ll p){
    for(int to:G[u]){
        if(to==p)continue;
        preDFS(to,u);
        siz[u]+=siz[to];
    }
    siz[u]++;
}
ll sum=1;
void DFS(ll u,ll p){
    ans[u]=sum;
    for(int to:G[u]){
        if(to==p)continue;
        ll a=siz[u],b=siz[to],c=sum;
        sum*=siz[to];
        sum%=mod;
        siz[u]-=siz[to];
        sum*=rui(siz[u],mod-2);
        sum%=mod;
        siz[to]=a;
        DFS(to,u);
        siz[u]=a;
        siz[to]=b;
        sum=c;
    }
}
int main(){
    ll n;cin>>n;
    rep(i,n-1){
        ll a,b;cin>>a>>b;
        a--;b--;
        G[a].pb(b);
        G[b].pb(a);
    }
    preDFS(0,-1);
    REP(i,1,n){
        sum*=i;
        sum%=mod;
    }
    REP(i,1,n){
        sum*=rui(siz[i],mod-2);
        sum%=mod;
    }
    DFS(0,-1);
    rep(i,n){
        cout<<ans[i]<<"\n";
    }
}