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;
#define lson root<<1,l,midd
#define rson root<<1|1,midd+1,r
#define pb push_back
typedef long long ll;
const int M=1e6+6;
const int mod=1e9+7;
vector<int>g[M];
ll facn=1;
int n;
ll ans[M],sz[M],mul[M],fac[M];
ll ksm(ll a,ll b){
    a%=mod;
    ll t=1;
    while(b){
        if(b&1)
            t=t*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return t;
}
void dfs1(int u,int fa){
    sz[u]=mul[u]=1;
    for(auto v:g[u]){
        if(v!=fa){
            dfs1(v,u);
            sz[u]+=sz[v];
            mul[u]=mul[u]*mul[v]%mod;
        }
    }
    mul[u]=mul[u]*sz[u]%mod;
}
void dfs2(int u,int fa,ll last){
    ll tmp=last;
    for(auto v:g[u])
        if(v!=fa)
            tmp*=mul[v],tmp%=mod;
    ans[u]=facn*ksm(tmp,mod-2)%mod;
    for(auto v:g[u])
        if(v!=fa)
            last*=mul[v],last%=mod;
    for(auto v:g[u])
        if(v!=fa)
            dfs2(v,u,last*ksm(mul[v],mod-2)%mod*(n-sz[v])%mod);
}
int main(){
    scanf("%d",&n);
    for(ll i=2;i<n;i++)
        facn*=i,facn%=mod;
    for(int u,v,i=1;i<n;i++){
        scanf("%d%d",&u,&v);
        g[u].pb(v);
        g[v].pb(u);
    }
    dfs1(1,0);
    dfs2(1,0,1);
    for(int i=1;i<=n;i++)
        printf("%lld\n",ans[i]);
    return 0;
}