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;

const int mod = 1000*1000*1000+7;
const int MAXN = 200*1000+5;

vector<int> g[MAXN];
int fact[MAXN], fact_inv[MAXN], ans[MAXN], sz[MAXN], n;

int mul(int a, int b) { 
	return 1LL * a * b % mod; 
}

int mpow(int a, int k) {
	if(k == 0) return 1;
	int x = mpow(a, k/2);
	x = mul(x, x);
	if(k % 2 == 1) x = mul(x, a);
	return x;
}

void dfs_pre(int u, int p = -1) {
	int& res = ans[u] = 1;
	sz[u] = 1;
	for(int v : g[u]) {
		if(v != p) {
			dfs_pre(v, u);
			res = mul(res, ans[v]);
			res = mul(res, fact_inv[sz[v]]);
			sz[u] += sz[v];
		}
	}
	res = mul(res, fact[sz[u]-1]);
}

void dfs(int u, int p = -1) {
	int& res = ans[u] = fact[n-1];
	for(int v : g[u]) {
		res = mul(res, ans[v]);
		res = mul(res, fact_inv[sz[v]]);
	}
	
	for(int v : g[u]) {
		if(v != p) {
			int tmp1 = ans[v], tmp2 = sz[u];
			res = mul(res, mpow(tmp1, mod-2));
			res = mul(res, fact_inv[n-1]);
			sz[u] = n - sz[v];
			res = mul(res, fact[sz[u]-1]);
			res = mul(res, fact[sz[v]]);
			dfs(v, u);
			res = mul(res, fact_inv[sz[v]]);
			res = mul(res, fact_inv[sz[u]-1]);
			sz[u] = tmp2;
			res = mul(res, fact[n-1]);
			res = mul(res, tmp1);
		}
	}
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	fact[0] = fact_inv[0] = 1;
	for(int i = 1; i < MAXN; i++) {
		fact[i] = mul(fact[i-1], i);
		fact_inv[i] = mpow(fact[i], mod-2);
	}
	
	cin >> n;
	for(int i = 0; i < n-1; i++) {
		int u, v;
		cin >> u >> v;
		u--, v--;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	
	dfs_pre(0);
	dfs(0);
	for(int i = 0; i < n; i++) {
		cout << ans[i] << '\n';
	}
}
