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], vr[2*MAXN], nx[2*MAXN], sz[MAXN], f[MAXN], tp=1;
void add_edge (int x,int y){
    vr[++eg]=y, nx[eg]=hd[x], hd[x]=eg;
}
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){
	sz[x]=1;
	for (register int i=hd[x]; i; i=nx[i]){
		if (vr[i]==fa) continue;
		dfs1 (vr[i],x);
		sz[x]+=sz[vr[i]];
	}
	f[1]=(1ll*f[1]*sz[x])%P;
}
void dfs2 (int x,int fa){
	for (register int i=hd[x]; i; i=nx[i]){
		if (vr[i]==fa) continue;
		f[vr[i]]=(1ll*f[x]*((1ll*(n-sz[vr[i]])*qpow(sz[vr[i]],P-2))%P))%P;
		dfs2 (vr[i],x);
	}
}
int main (){
	ios::sync_with_stdio(false);
	cin >> n;
	f[1]=1;
	for (register int i=1; i<n; i++){
		cin >> x >> y;
		add_edge(x,y), add_edge(y,x);
	}
	dfs1(1,0), dfs2(1,0);
	for (register int i=1; i<=n; i++) tp=(1ll*tp*i)%P;
	for (register int i=1; i<=n; i++) cout << (1ll*tp*qpow(f[i],P-2))%P << endl;
	return 0;
}