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;
struct mint{
	static constexpr int mod = 1e9+7;
	int x;
	mint(ll x=0): x((x %= mod) < 0 ? x+mod : x){}
	mint operator-(){return mint(mod-x);}
	mint &operator+=(const mint &n){if((x += n.x) >= mod) x -= mod; return *this;}
	mint &operator-=(const mint &n){if((x += n.x) >= mod) x -= mod; return *this;}
	mint &operator*=(const mint &n){x = (ll)x * n.x % mod; return *this;}
	mint &operator/=(const mint &n){return *this *= n.inv();}
	mint &operator^=(const mint &n){
		mint ret{1}, tmp{x};
		for(int t = n.x; t; t >>= 1){
			if(t&1) ret *= tmp;
			tmp *= tmp;
		}
		return *this = ret;
	}
	mint inv() const{return *this ^ (mod - 2);}
	mint operator+(const mint &n){return mint(*this) += n;}
	mint operator-(const mint &n){return mint(*this) -= n;}
	mint operator*(const mint &n){return mint(*this) *= n;}
	mint operator/(const mint &n){return mint(*this) /= n;}
	mint operator^(const mint &n) const{return mint(*this) ^= n;}
};
constexpr int MX = 200005;
int n, v, u, s[MX];
vector<int> G[MX];
mint ans[MX], ac = 1;
mint dfs(int u, int p){
	mint v{1};
	s[u] = 1;
	for(int nx: G[u]) if(nx != p){
		v *= dfs(nx, u);
		s[u] += s[nx];
	}
	v /= s[u];
	return v;
}
void dfs2(int u, int p){
	if(u != p) ans[u] = ans[p] * s[u] / (n-s[u]);
	for(int nx: G[u]) if(nx != p) dfs2(nx, u);
}
int main(){
	scanf("%d", &n);
	rep(i,n) ac *= i+1;
	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;
	dfs2(0,0);
	rep(i,n) printf("%d\n", ans[i].x);
}