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 rep(i,n) for(int i=0; i<(n); i++)
using ll = long long;
constexpr int MOD = 1e9+7, MX = 200005;
int n, v, u, s[MX];
vector<int> G[MX];
ll ans[MX], ac = 1;

ll rev(ll a){
	ll b = MOD, x = 1, y = 0, u = 0, v = 1;
	while(b){
		ll t = a / b;
		swap(a -= t * b, b);
		swap(x -= t * u, u);
		swap(y -= t * v, v);
	}
	x %= MOD;
	return x < 0 ? x+MOD : x;
}
ll dfs(int u, int p){
	ll v = s[u] = 1;
	for(int nx: G[u]) if(nx != p){
		(v *= dfs(nx, u)) %= MOD;
		s[u] += s[nx];
	}
	(v *= rev(s[u])) %= MOD;
	return v;
}
void dfs2(int u, int p){
	if(u != p) ans[u] = ans[p] * s[u] % MOD * rev(n-s[u]) % MOD;
	for(int nx: G[u]) if(nx != p) dfs2(nx, u);
}
int main(){
	scanf("%d", &n);
	rep(i,n) (ac *= i+1) %= MOD;
	rep(i,n-1){
		scanf("%d%d", &v, &u);
		v--; u--;
		G[v].push_back(u);
		G[u].push_back(v);
	}
	ans[0] = dfs(0,0) * ac % MOD;
	dfs2(0,0);
	rep(i,n) printf("%lld\n", ans[i]);
}