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>
#define mod 1000000007
#define m_k make_pair
#define inf 0x3f3f3f3f
#define int long long
using namespace std;
const int N=2*1e5+100;
int n,dp[N],sz[N];
vector <int> e[N];
int m_pow(int a,int b)
{
	int ans=1;
	while (b)
	{
		if (b&1) ans=(ans*a)%mod;
		b>>=1;
		a=(a*a)%mod;
	}
	return ans;
}
int inv(int x)
{
	return m_pow(x,mod-2);
}
void dfs(int x,int fa)
{
	sz[x]=1;
	for (int i=0;i<(int)e[x].size();i++)
	{
		int u=e[x][i];
		if (u==fa) continue;
		dfs(u,x);
		sz[x]+=sz[u];
	}
}
void dfs1(int x,int fa)
{
	for (int i=0;i<(int)e[x].size();i++)
	{
		int u=e[x][i];
		if (u==fa) continue;
		dp[u]=(dp[x]*(n-sz[u]))%mod;
		dp[u]=(dp[u]*inv(sz[u]))%mod;
		dfs1(u,x);
	}
}
signed main()
{
	scanf("%lld",&n);
	for (int i=1;i<n;i++)
	{
		int a,b;
		scanf("%lld%lld",&a,&b);
		e[a].push_back(b);
		e[b].push_back(a);
	}
	dfs(1,1);
	dp[1]=1;
	for (int i=1;i<=n;i++) dp[1]=(dp[1]*sz[i])%mod;
	dfs1(1,1);
	int ans=1;
	for (int i=1;i<=n;i++) ans=(ans*i)%mod;
	for (int i=1;i<=n;i++) dp[i]=(ans*inv(dp[i]))%mod;
	for (int i=1;i<=n;i++) printf("%lld\n",dp[i]);
	printf("\n");
}