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;
#define mp(a, b) make_pair(a, b)
#define ls (rt << 1)
#define rs (rt << 1 | 1) 
typedef pair<int, int> pii; 
const ll mod = 1e9 + 7;
template <class T> T lowbit(T x){return ((x)&(-(x)));}
const int maxn = 2e5 + 10;
int n;
vector<int> E[maxn];
ll fac[maxn], inv[maxn], facinv[maxn], bin[maxn];
int siz[maxn];
ll f[maxn];
void pre()
{
	fac[0] = 1; for(int i = 1; i < maxn; ++i) fac[i] = fac[i - 1] * i % mod;
	inv[1] = 1; for(int i = 2; i < maxn; ++i) inv[i] = (mod - mod / i) * inv[mod % i] % mod;
	facinv[0] = 1; for(int i = 1; i < maxn; ++i) facinv[i] = facinv[i - 1] * inv[i] % mod;
	bin[0] = 1; for(int i = 1; i < maxn; ++i) bin[i] = bin[i - 1] * 2 % mod;
}
void dfs1(int x, int fa)
{
	siz[x] = 1;
	for(auto v : E[x])
		if(v != fa)
		{
			dfs1(v, x);
			siz[x] += siz[v];
			f[1] = f[1] * inv[siz[v]] % mod;
		}
}
void dfs2(int x, int fa)
{
	if(x != 1)
	{
		f[x] = f[fa] * siz[x] % mod * inv[n - siz[x]] % mod;
	}
	for(auto v : E[x])
		if(v != fa)
		{
			dfs2(v, x);
		}
}
int main()
{
	pre();
	scanf("%d", &n);
	for(int i = 1; i < n; ++i) 
	{
		int u, v;
		scanf("%d%d", &u, &v);
		E[u].push_back(v);
		E[v].push_back(u);
	}
	f[1] = fac[n - 1];
	dfs1(1, 0);
	dfs2(1, 0);
	for(int i = 1; i <= n; ++i)
		printf("%lld\n", f[i]);
	return 0;
}