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;
const LL MOD = 1e9 + 7;
const int N = 200000 + 10;
int n, sz[N]; 
vector<int> g[N];
LL fac[N];
LL mpow(LL a, LL x){
	if (x==0) return 1;
	LL t = mpow(a, x>>1);
	if (x%2==0) return t*t%MOD;
	return t*t%MOD*a%MOD;
}
void dfs1(int u, int p) {
	sz[u] = 1;
	for (auto v: g[u]) {
		if (v == p) continue;
		dfs1(v, u);
		sz[u] += sz[v];
	}
}
LL ans[N];
void dfs2(int u, int p) {
	for (auto v: g[u]) {
		if (v == p) continue;
		ans[v] = ans[u] * sz[v] % MOD;
		ans[v] = ans[v] * mpow(n - sz[v], MOD - 2) % MOD;
		dfs2(v, u);
	}
}
int main() {
	fac[0] = 1;
	for (int i = 1; i < N; i ++) fac[i] = fac[i-1] * i % MOD;
	scanf("%d",&n);
	for(int i=1;i<n;i++){
		int u,v; scanf("%d%d",&u,&v);
		g[u].push_back(v);
		g[v].push_back(u);
	}
	dfs1(1, 1);
	ans[1] = fac[n];
	for (int i = 1; i <= n; i ++)
		ans[1] = ans[1] * mpow(sz[i], MOD-2) % MOD;
	dfs2(1, 1);
	for (int i = 1; i <= n; i ++)
		printf("%lld\n", ans[i]);
}