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;
#define ll long long
const ll mod = 1e9 + 7;
const int N = 2e5 + 10;
int n;
struct Edge {
	int to, nxt;
}e[N << 1];
int tot, head[N];
ll siz[N],son[N], sum = 1;
ll ans[N];
ll qpow(ll x, ll k) {
	ll res = 1;
	while (k) {
		if (k & 1) res = res * x % mod;
		k >>= 1;
		x = x * x % mod;
	}
	return res;
}
void add(int x, int y) {
	e[++tot] = Edge{ y,head[x] };
	head[x] = tot;
}
void predfs(int x, int fa) {
	siz[x] = son[x] = 1;
	for (int i = head[x]; i; i = e[i].nxt) {
		int to = e[i].to;
		if (to == fa)
			continue;
		predfs(to, x);
		son[x] += son[to];
		siz[x] = siz[x] * siz[to] % mod;
	}
	siz[x] = siz[x] * son[x] % mod;
}
void dfs(int x,int fa,ll pre) {
	ll t = pre;
	for (int i = head[x]; i; i = e[i].nxt) {
		int to = e[i].to;
		if (to == fa)
			continue;
		t = t * siz[to] % mod;
	}
	ans[x] = sum * qpow(t, mod - 2) % mod;
	for (int i = head[x]; i; i = e[i].nxt) {
		int to = e[i].to;
		if (to == fa)
			continue;
		pre = pre * siz[to] % mod;
	}
	for (int i = head[x]; i; i = e[i].nxt) {
		int to = e[i].to;
		if (to == fa)
			continue;
		dfs(to, x, pre * qpow(siz[to], mod - 2) % mod * (n - son[to]) % mod);
	}
}
int main() {
	scanf("%d", &n);
	for (int i = 1; i < n; i++)
	{
		int x, y;
		scanf("%d %d", &x, &y);
		add(x, y), add(y, x);
		sum = sum * i % mod;
	}
	predfs(1, 0);
	dfs(1, 0, 1);
	for (int i = 1; i <= n; i++)
		printf("%lld\n", ans[i]);
}