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 Mod 1000000007
#define Val(x) (((long long)x) % Mod)
#define maxn 200005
using namespace std;
int n;
vector<int> G[maxn];
int fac[maxn],sz[maxn],dp[maxn];
int Pow(int base,int n) {
	int ret = 1;
	while(n > 0) {
		if(n & 1)	ret = Val(ret * base);
		base = Val(base * base);
		n >>= 1;
	}
	return ret;
}
int inv(int val) {
	return Pow(val,Mod-2);
}
void dfs1(int u,int pa) {
	dp[u] = sz[u] = 1;
	for(int v : G[u])
		if(v != pa) {
			dfs1(v,u);
			sz[u] += sz[v];
			dp[u] = Val(dp[u] * dp[v]);
			dp[u] = Val(dp[u] * inv(fac[sz[v]]));
		}
	dp[u] = Val(dp[u] * fac[sz[u] - 1]);
}
int ans[maxn];
void dfs2(int u,int pa,int up) {
	dp[u] = Val(dp[u] * inv(fac[sz[u] - 1]));
	ans[u] = Val(Val(dp[u] * up) * fac[n - 1]);
	for(int v : G[u])
		if(v != pa) {
			int down = Val(Val(dp[u] * inv(dp[v])) * fac[sz[v]]);
			down = Val(down * up);
			down = Val(Val(down * fac[n - sz[v] - 1]) * inv(fac[n - sz[v]]));
			dfs2(v,u,down);
		}
}
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	fac[0] = 1;
	for(int i=1;i<maxn;i++)
		fac[i] = Val(fac[i-1] * i);
	cin >> n;
	for(int i=1,u,v;i<n;i++) {
		cin >> u >> v;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	dfs1(1,1);
	dfs2(1,1,1);
	for(int i=1;i<=n;i++)
		cout << ans[i] << endl;
	return 0;
}
