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;
const int N = 2e5 + 5, MOD = 1e9 + 7;
struct edge
{
	int v, nxt;
}e[N * 2];
int p[N], eid;
int res[N],sz[N];
bool vst[N];
int n;

void add(int u, int v)
{
	e[eid] = edge{v, p[u]}; p[u] = eid ++;
}
int pow_mod(int a, int b)
{
	long long res = 1, tmp = a;
	for (; b; b /= 2)
	{
		if(b & 1) res = res * tmp % MOD;
		tmp = tmp * tmp % MOD;
	}
	return res;
}
void dfs1(int u)
{
	vst[u] = 1;
	sz[u] = 1;
	
	for (int i = p[u]; ~i; i = e[i].nxt)
	{
		int v = e[i].v;
		if(vst[v]) continue;
		dfs1(v);
		sz[u] += sz[v];
	}
	res[1] = (1LL * res[1] * u) % MOD;
	res[1] = (1LL * res[1] * pow_mod(sz[u], MOD - 2)) % MOD;
}
void dfs2(int u)
{
	vst[u] = 1;
	for (int i = p[u]; ~i; i = e[i].nxt)
	{
		int v = e[i].v;
		if(vst[v]) continue;
		res[v] = 1LL * res[u] * sz[v] % MOD * pow_mod(n - sz[v], MOD - 2) % MOD;
		dfs2(v);
	}
}
int main()
{
	scanf("%d", &n);
	memset(p, -1, sizeof(p));
	for (int i = 1; i < n; ++ i)
	{
		int u ,v; scanf("%d%d", &u, &v);
		add(u, v);
		add(v, u);
	}
	res[1] = 1;
	dfs1(1);
	memset(vst, 0, sizeof(vst));
	dfs2(1);
	for (int i = 1; i <= n; ++ i) printf("%d\n", res[i]);
	return 0;
}