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 mp make_pair
#define ll long long
#define x first
#define y second
int mod = 1000 * 1000 * 1000 + 7;
const int MAX = 200005;
int fact[200005];
int inv[200005];
int add(int a , int b)
{
	return a + b >= mod ? a + b - mod : a + b;
}
int mult(int a, int b)
{
	return a * (ll)b % mod;
}
int modPow(int a, int step)
{
	int ans = 1;
	while(step)
	{
		if(step & 1)
		{
			ans = mult(ans , a);
		}
		a = mult(a , a);
		step >>= 1;
	}
	return ans;
}
int c(int n , int k)
{
	return mult(fact[n] , mult(inv[k] , inv[n - k]));
}
vector<int> g[200005];
int sub[200005];
int f[200005];
int dp[200005];
int n;
void dfs(int u, int p = -1)
{
	sub[u] = 0;
	f[u] = 1;
	for(auto v : g[u])
	{
		if(v != p)
		{
			dfs(v, u);
			sub[u] += sub[v];
			f[u] = mult(mult(f[u] , f[v]) , c(sub[u] , sub[v]));
		}
	}
	sub[u]++;
}
void dfs2(int u, int fP , int p = -1)
{
	dp[u] = mult(mult(f[u] , fP) , c(n - 1 , sub[u] - 1));
	for(auto v : g[u])
	{
		if(v == p)
			continue;
		int fS = mult(mult(f[u] , modPow(f[v] , mod - 2)) , modPow(c(sub[u] - 1 , sub[v]) , mod - 2));
		dfs2(v, mult(mult(fP , fS) , c(n - sub[v] - 1 , n - sub[u])) ,  u);
	}
}
int main()
{
	ios_base::sync_with_stdio(0);
	fact[0] = inv[0] = 1;
	for(int i = 1; i < MAX; i++)
		fact[i] = mult(fact[i - 1] , i);
	for(int i = 1; i < MAX; i++)
		inv[i] = modPow(fact[i] , mod - 2);
	
	cin >> n;
	for(int i = 1; i < n; i++)
	{
		int u , v;
		cin >> u >> v;
		u--;v--;
		g[u].pb(v);
		g[v].pb(u);
	}
	dfs(0);
	dfs2(0 , 1);
	for(int i = 0; i < n; i++)
		cout << dp[i] << endl;
}
