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 MAXN 200010
#define ll long long

using namespace std;

const int mod = 1e9+7;

int n, tot;

ll product = 1;

int head[MAXN], siz[MAXN];

ll f[MAXN];

struct edge {
	int to, next;
}e[MAXN * 2];

void add_edge(int x, int y) {
	e[++tot].to = y;
	e[tot].next = head[x];
	head[x] = tot;
	return;
}

ll qp(ll a, ll b) {
	ll res = 1, base = a;
	while (b) {
		if (b & 1) res = (res * base) % mod;
		base = (base * base) % mod;
		b >>= 1;
	}
	
	return res;
}

void dfs1(int x, int fa) {
	siz[x] = 1;
	for (int i = head[x]; i; i = e[i].next) {
		int v = e[i].to;
		if (v == fa) continue;
		dfs1(v, x);
		siz[x] += siz[v];
	}
	
	f[1] = (1ll * f[1] * siz[x]) % mod;
	return;
}

void dfs2(int x, int fa) {
	for (int i = head[x]; i; i = e[i].next) {
		int v = e[i].to;
		if (v == fa) continue;
		f[v] = (1ll * f[x] * (1ll * (n - siz[v]) * qp(siz[v], mod - 2) % mod)) % mod;
		dfs2(v, x);
	}
	
	return;
}

int main() {
	scanf("%d", &n);
	for (int i = 1, u, v; i < n; i++) {
		scanf("%d%d", &u, &v);
		add_edge(u, v); add_edge(v, u);
	}
	
	f[1] = 1;
	dfs1(1, 0);
	dfs2(1, 0);
	
	for (int i = 1; i <= n; i++) {product = (1ll * product * i) % mod;}
	for (int i = 1; i <= n; i++) {printf("%lld\n", (ll)(1ll * product * qp(f[i], mod - 2)) % mod);}
	return 0;
}