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 <cstdio>
#include <vector>
#define N 200005
using namespace std;
typedef long long ll;

ll n, nfac=1, sz[N], num[N]={0,1}, M=1e9+7;
vector<ll> gr[N];

ll pow(ll p, ll q) {
	ll re = 1;
	while(q) {
		if(q&1) re = re * p % M;
		p = p * p % M;
		q >>= 1;
	}
	return re;
}

void dfs(int u, int pr) {
	sz[u] = 1;
	for(auto v:gr[u]) {
		if(v==pr) continue;
		dfs(v, u);
		sz[u] += sz[v];
	}
	(num[1] *= sz[u]) %= M;
}

void dfs2(int u, int pr) {
	for(auto v:gr[u]) {
		if(v==pr) continue;
		num[v] = num[u] * pow(sz[v], M-2) % M * (n-sz[v]) % M;
		dfs2(v, u);
	}
}

int main() {
	ll i, t1, t2;
	cin >> n;
	for(i=0; i<n-1; i++) {
		scanf("%lld%lld", &t1, &t2);
		gr[t1].push_back(t2);
		gr[t2].push_back(t1);
	}
	for(i=2; i<=n; i++)	(nfac *= i) %= M;
	dfs(1, 0);
	dfs2(1, 0);
	for(i=1; i<=n; i++) {
		printf("%lld\n", nfac * pow(num[i], M-2) % M);
	}
	return 0;
}