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 pb push_back
#define F first
#define S second
#define mkp make_pair
#define pii pair<int,int>
typedef long long ll;
const int inf=0x3f3f3f3f;
const int maxn=2e5+5;
const int mod=1e9+7;
int fac[maxn],inv[maxn],dp[maxn],sz[maxn];
int n;
vector<int>v[maxn];
ll mul(ll a,ll b){
	return (a*b)%mod;
}
void init(int n){
	fac[1]=1;
	fac[0]=1;
	for(int i=2;i<=n;i++){
		fac[i]=mul(fac[i-1],i);
	}
	inv[1] = 1;
	for(int i = 2; i <=n; ++ i)
    inv[i] = mul((mod - mod / i) , inv[mod % i] );
}
void dfs(int x,int fa){
	sz[x]=1;
	for(auto y:v[x]){
		if(y==fa)continue;
		dfs(y,x);
		sz[x]+=sz[y];
	}
}
void reroot(int x,int fa){
	dp[x]=mul(mul(dp[fa],sz[x]),inv[(n-sz[x])]);
	for(auto p:v[x]){
		if(p==fa)continue;
		reroot(p,x);
	}
}
int main(){
	
	scanf("%d",&n);
	init(n);
	for(int i=1;i<n;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		v[x].pb(y);
		v[y].pb(x);
	}
	dfs(1,-1);
	dp[1]=fac[n];
	for(int i=1;i<=n;i++){
		dp[1]=mul(dp[1],inv[sz[i]]);
	}

	for(auto p:v[1]){
		reroot(p,1);
	}
	for(int i=1;i<=n;i++)printf("%d\n",dp[i]);
	return 0;
}
