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=200010,P=1e9+7;
int n,x,y,eg,hd[MAXN],ver[2*MAXN],nx[2*MAXN],siz[MAXN],f[MAXN];
void add_edge (int x,int y) {
    ver[++eg]=y;
    nx[eg]=hd[x];
    hd[x]=eg;
    return;
}
int qpow (int a,int b) {
    int res=1;
    while (b) {
        if (b&1) {res=(1ll*res*a)%P;}
        a=(1ll*a*a)%P,b>>=1; 
    }
    return res;
}
void dfs1 (int x,int fa) {
    siz[x]=1;
    for (int i=hd[x];i;i=nx[i]) {
        if (ver[i]==fa) {continue;}
        dfs1(ver[i],x);
        siz[x]+=siz[ver[i]];
    }
    f[1]=(1ll*f[1]*siz[x])%P;
    return;
}
void dfs2 (int x,int fa) {
    for (int i=hd[x];i;i=nx[i]) {
        if (ver[i]==fa) {continue;}
        f[ver[i]]=(1ll*f[x]*((1ll*(n-siz[ver[i]])*qpow(siz[ver[i]],P-2))%P))%P; 
        dfs2(ver[i],x);
    }
    return;
}
int main () {
    scanf("%d",&n);
    f[1]=1;
    for (int i=1;i<=n-1;i++) {
        scanf("%d%d",&x,&y);
        add_edge(x,y),add_edge(y,x);
    }
    dfs1(1,0);
    dfs2(1,0);
    int tmp=1;
    for (int i=1;i<=n;i++) {tmp=(1ll*tmp*i)%P;}
    for (int i=1;i<=n;i++) {printf("%d\n",(1ll*tmp*qpow(f[i],P-2))%P);}
    return 0;
}