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<iostream>
#include <algorithm>
#include<queue>
#include<cmath>
#include<math.h>
#include<string>
#include<string.h>
#include<map>
#include<set>
using LL = long long;
using namespace std;

const int M = 2e5 + 100;
const int mod = 1e9 + 7;

vector<int> e[M];
long long n, sz[M], ans[M];
long long fac[M], inv[M];

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

void init() {
	fac[0] = 1;
	for (int i = 1; i <= n; i++)
		fac[i] = mul(fac[i - 1], i);
	inv[1] = 1;
	for (int i = 2; i <= n; i++)
		inv[i] = mul((mod - mod / i), inv[mod % i]);
}

void dfs(int u, int fa) {
	sz[u] = 1;
	for (int v : e[u]) {
		if (v != fa) {
			dfs(v, u);
			sz[u] += sz[v];
		}
	}
}

void reroot(int u, int fa) {
	ans[u] = mul(mul(ans[fa], sz[u]), inv[n - sz[u]]);
	for (int v : e[u]) {
		if (v != fa) {
			reroot(v, u);
		}
	}
}

int main() {
	cin >> n;
	init();
	for (int i = 0; i < n - 1; i++) {
		int u, v; cin >> u >> v;
		e[u].push_back(v);
		e[v].push_back(u);
	}
	dfs(1, 0);
	ans[1] = fac[n];
	for (int i = 1; i <= n; i++)
		ans[1] =  ans[1] *inv[ sz[i]] % mod;
	for (int v : e[1]) reroot(v, 1);
	for (int i = 1; i <= n; i++) cout << ans[i] << "\n";
}
