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<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N=2e5+10,mod=1e9+7;
int first[N],n,siz[N],ans[N],tot,mul[N],fac[N];
int pew(int x,int b){
    int res=1;
    while(b){
        if(b%2==1){
            res=1ll*res*x%mod;
        }
        b/=2;
        x=1ll*x*x%mod;
    }
    return res;
}
struct edge{
    int to,next;
}e[N*2];
void add(int u,int v){
    e[++tot].next=first[u],first[u]=tot;
    e[tot].to=v;
}
void dfs(int x,int f){
    siz[x]=1;mul[x]=1;
    for(int i=first[x];i;i=e[i].next){
        int u=e[i].to;
        if(u==f)continue;
        dfs(u,x);siz[x]+=siz[u];
        mul[x]=1ll*mul[x]*mul[u]%mod;
    }
    mul[x]=1ll*mul[x]*siz[x]%mod;
}
void getans(int x,int f,int now){
    int down=now;
    for(int i=first[x];i;i=e[i].next){
        int u=e[i].to;
        if(u==f)continue;
        down=1ll*down*mul[u]%mod;
        now=1ll*now*mul[u]%mod;
    }
    ans[x]=1ll*fac[n-1]*pew(down,mod-2)%mod;
    for(int i=first[x];i;i=e[i].next){
        int u=e[i].to;
        if(u==f)continue;
        getans(u,x,1ll*now*pew(mul[u],mod-2)%mod*(n-siz[u])%mod);
    }
}
int main(){
    scanf("%d",&n);
    for(int i=1,u,v;i<n;i++){
        scanf("%d%d",&u,&v);
        add(u,v);add(v,u);
    }
    fac[0]=1;
    for(int i=1;i<=n;i++)
        fac[i]=1ll*fac[i-1]*i%mod;
    dfs(1,0);
    getans(1,0,1);
    for(int i=1;i<=n;i++)
        printf("%d\n",ans[i]);
}
