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 mp make_pair
#define pb push_back
#define MOD 1000000007
typedef long long ll;
int n;
vector<vector<int>> v(200005);
bool visit[200005];
int d[200005], child_num[200005];
ll f[200005], ans[200005];

int dfs(int start, int depth)
{
	int ret = 0;
	visit[start] = true;
	d[start] = depth;
	for(int i : v[start]){
		if(!visit[i]) {
			ret += dfs(i, depth+1);
		}
	}
	return child_num[start] = ret + 1;
}

ll power(ll a, ll n) {
	ll res = 1;
	while(n > 0) {
		if(n & 1) {
			res *= a;
			res %= MOD;
		}
		a *= a;
		a %= MOD;
		n >>= 1;
	}
	return res;
}

void get_ans(int start)
{
	if(child_num[start]==1) return;
	ans[1] = (ans[1] * (f[child_num[start] -1])) % MOD;
	for(int i : v[start]) {
		if(d[i] < d[start]) continue;
		ans[1] *= power(f[child_num[i]], MOD - 2);
		ans[1] %= MOD;
		get_ans(i);
	}
}

void dfs2(int start)
{
	visit[start] = true;
	for(int i : v[start]){
		if(!visit[i]) {
			ans[i] = (ans[start] * f[child_num[i]] ) % MOD;
			ans[i] = (ans[i] * power(f[n - child_num[i]], MOD - 2))  % MOD;
			ans[i] = (ans[i] * f[n - child_num[i]-1])  % MOD;
			ans[i] = (ans[i] * power(f[child_num[i] -1], MOD - 2)) % MOD;
			dfs2(i);
		}
	}
}

int main(void)
{
	cin >> n;
	f[0] = 1; ans[1] = 1;
	for(int i=1;i<=n;i++) {
		f[i] = (f[i-1] * i) % MOD;
	}
	int from, to;
	for(int i=1;i<n;i++){
		scanf("%d %d", &from, &to);
		v[from].pb(to);
		v[to].pb(from);
	}
	dfs(1, 0);
	get_ans(1);
	for(int i=1;i<=n;i++) visit[i] = false;
	dfs2(1);
	for(int i=1;i<=n;i++) cout << ans[i] << endl;
	return 0;
}
