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;
typedef long long ll;
const int N=2e5+200,mod=1e9+7;
vector<int> g[N];
int n;

ll sz[N],tmp=1,a[N];
ll qpow(ll a,ll b)
{
	ll ans=1;
	while(b){
		if(b&1)ans=(ans*a)%mod;
		a=a*a%mod;
		b>>=1;
	}
	return ans;
}
void dfs(int u,int fa)
{
	sz[u]=1;
	for(auto to:g[u])
	{
		if(to==fa)continue;
		dfs(to,u);
		sz[u]+=sz[to];
	}
	tmp=(tmp*sz[u])%mod;
}
void dfs2(int u,int fa)
{
	for(auto to:g[u])
	{
		if(to==fa)continue;
		a[to]=a[u]*qpow(sz[to],mod-2)%mod*(n-sz[to])%mod;
		dfs2(to,u);
	}
}
int main()
{
    scanf("%d",&n);
    ll lc=1;
    for(ll i=2;i<=n;i++)
    {
    	int a,b;
		scanf("%d%d",&a,&b);
    	g[a].push_back(b);
    	g[b].push_back(a);
    	lc=(lc*i)%mod;
	}
	dfs(1,-1);
	a[1]=tmp;
	dfs2(1,-1);
	for(int i=1;i<=n;i++)printf("%lld\n",lc*qpow(a[i],mod-2)%mod);
} 