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<iostream>
#include<algorithm>
#include<queue>
#include<cmath>
#include<math.h>
#include<string>
#include<string.h>
#include<map>
#include<unordered_map>
#include<unordered_set>
#include<set>
#include<stack>
#include<sstream>
//#include <Eigen/Dense>
//#include <Eigen/Geometry>
//using namespace Eigen;
using namespace std;

int n;
const int maxn = 2e5 + 100;
int mod = 1e9 + 7;
vector<int> g[maxn];
int vis[maxn], sz[maxn];
long long ans[maxn], inv[maxn];

void inv_n(int n)
{
	inv[1] = 1;
	for (long long i = 2; i <= n; i++)
	{
		inv[i] = inv[mod%i] * (mod - mod / i) % mod;
	}
}


long long dfs1(int cur,int p)
{
	sz[cur] = 1;
	for (int i : g[cur])
	{
		if (i != p)	
			sz[cur] += dfs1(i, cur);
	}
	return sz[cur];
}

void dfs2(int cur, int p)
{
	ans[cur] = ans[p] * sz[cur] % mod * inv[(n - sz[cur])] % mod;//ans[p] * sz[cur] * inv[(n - sz[cur])] % mod溢出报错
	for (int i : g[cur])
	{
		if (i != p)
			dfs2(i,cur);
	}
}
int main()
{
	int u, v;
	cin >> n;
	inv_n(n);
	for (int i = 0; i < n - 1; i++)
	{
		cin >> u >> v;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	ans[1] = 1;
	for (int i = 1; i <= n; i++)
		ans[1] = ans[1] * i%mod;
	dfs1(1, 0);
	for (int i = 1; i <= n; i++)
		ans[1] = ans[1] * inv[sz[i]] % mod;
	
	for (int i : g[1])
		dfs2(i, 1);
	for (int i = 1; i <= n; i++)
		cout << ans[i] << endl;

	return 0;
}