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>
#define int long long
#define MOD 1000000007
#define N (int) 3e5
using namespace std;
int n;
struct Edge {
	int v, nxt;
} edge[N<<1];
int fst[N], ecnt;
void add_edge (int u, int v) {
	edge[ecnt].v = v, edge[ecnt].nxt = fst[u], fst[u] = ecnt ++;
}
void init_e () {
	memset (fst, -1, sizeof (fst));
}

int ksm (int x, int y) {
	int ret = 1;
	while (y) {
		if (y & 1) (ret *= x) %= MOD;
		(x *= x) %= MOD;
		y >>= 1;
	}
	return ret;
}

int inv (int x) {
	return ksm (x, MOD-2);
}

int sz[N], dp1[N], dp2[N];
void DP1 (int u, int f) {
	sz[u] = 1;
	dp1[u] = 1;
	for (int e = fst[u]; e != -1; e = edge[e].nxt) {
		int v = edge[e].v;
		if (v == f) continue;
		DP1 (v, u);
		sz[u] += sz[v];
		(dp1[u] *= dp1[v]) %= MOD;	
	}
	(dp1[u] *= sz[u]) %= MOD;
//	cout << u << " " << dp1[u] << endl;
}
void DP2 (int u, int f) {
	if (u != 1) dp2[u] = dp2[f] * (n - sz[u]) % MOD * inv (sz[u]) % MOD;
//	cout << u << " " << dp2[u] << endl;
	for (int e = fst[u]; e != -1; e = edge[e].nxt) {
		int v = edge[e].v;
		if (v == f) continue;
		DP2 (v, u);
	}
}

signed main () {
	init_e ();
	
	cin >> n;
	for (int i = 1; i < n; i ++) {
		int u, v; cin >> u >> v;
		add_edge (u, v); add_edge (v, u);
	}
	
	DP1 (1, 0);
	dp2[1] = dp1[1];
	DP2 (1, 0);
	
	int tmp = 1;
	for (int i = 1; i <= n; i ++) (tmp *= i) %= MOD;
	
	for (int i = 1; i <= n; i ++) cout << (1ll * tmp * inv (dp2[i]) % MOD) << endl;
	
	return 0;
}