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"
#define fio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define endl '\n'
#define all(V) (V).begin(), (V).end() 
using namespace std;
typedef long long ll;

ll mod = 1e9 + 7;

ll power(ll base, ll exponent, ll modx = mod) {
    ll ret = 1 % modx; 
    base %= modx; 
    while(exponent) {
        if(exponent & 1) {
            ret = (ret * base) % modx;
        }
        base = (base * base) % modx;
        exponent >>= 1;
    }
    return ret;
}
ll inv(ll val, ll modx = mod) {
    return power(val, modx - 2, modx);
}
vector <ll> ans, sub, sub_val;
vector <vector <int> > T;

void dfs(int u, int par) {
	sub[u] = 1;
	for(auto v : T[u]) {
		if(v != par) {
			dfs(v, u);
			sub[u] += sub[v];
			sub_val[u] = (sub_val[u] * sub_val[v]) % mod;
		}
	}
	sub_val[u] = (sub_val[u] * sub[u]) % mod;
}
int N;

void dfs2(int u, int par) {
	if(par != -1) {
		ans[u] = (ans[par] * inv(sub[u])) % mod;
		ans[u] = (ans[u] * (N - sub[u])) % mod;
	} else {
		ans[u] = sub_val[u];
	}
	for(auto v : T[u]) {
		if(v != par) {
			dfs2(v, u);
		}
	}

}

int main() {fio;
	cin >> N;
	T.resize(N);	
	ans.resize(N);
	sub.resize(N);
	sub_val.resize(N, 1);
	for(int i = 0; i < N - 1; i++) {
		int u, v;
		cin >> u >> v;
		u--; v--;
		T[u].push_back(v);
		T[v].push_back(u);
	}
	dfs(0, -1);
	dfs2(0, -1);
	ll nf = 1;
	for(int i = 1; i <= N; i++) {
		nf = (nf * i) % mod;
	}
	for(int i = 0; i < N; i++) {
		// cout << ans[i] << " -> ";
		cout << (inv(ans[i]) * nf) % mod << endl;
	}
    return 0;
}