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;

typedef long long ll;
const int maxn = 2E+5 + 5;
const ll mod = 1E+9 + 7;

int n;
ll siz[maxn], fa[maxn], dp[maxn], ans[maxn];
ll fac[maxn], inv[maxn];
vector<int> to[maxn];

ll qPow(ll a, ll b)
{
	a %= mod;
	
	ll res = 1;
	while(b) {
		if(b & 1) res = res * a % mod;
		a = a * a % mod, b >>= 1;
	}
	return res;
}

void DFS(int u, int father)
{
	fa[u] = father, siz[u] = 1, dp[u] = 1;
	for(int i = 0; i < to[u].size(); ++i) {
		int v = to[u][i];
		
		if(v == father) continue;
		
		DFS(v, u), siz[u] += siz[v], dp[u] = dp[u] * dp[v] % mod * qPow(siz[v], mod - 2) % mod;
	}
}

void Work(int u, ll tmp)
{
	for(int i = 0; i < to[u].size(); ++i) {
		int v = to[u][i];
		
		if(v == fa[u]) continue;
		
		Work(v, tmp * dp[u] % mod * qPow((n - siz[v]) * dp[v], mod - 2) % mod * siz[v] % mod);
	}
	ans[u] = tmp * dp[u] % mod * fac[n - 1] % mod;
}

int main()
{
	int N = 2E+5;
	
	fac[0] = 1;
	for(int i = 1; i <= N; ++i)
		fac[i] = fac[i - 1] * i % mod;
	inv[N] = qPow(fac[N], mod - 2);
	for(int i = N; i >= 1; --i)
		inv[i - 1] = inv[i] * i % mod;
	
	scanf("%d", &n);
	for(int i = 1; i < n; ++i) {
		int x, y;
		
		scanf("%d%d", &x, &y);
		to[x].push_back(y), to[y].push_back(x);
	}
	
	DFS(1, 0), ans[1] = dp[1] * fac[n - 1] % mod;
	for(int i = 0; i < to[1].size(); ++i) {
		int v = to[1][i];
		
		Work(v, dp[1] * qPow((n - siz[v]) * dp[v], mod - 2) % mod * siz[v] % mod);
	}
	
	for(int i = 1; i <= n; ++i)
		printf("%lld\n", ans[i]);
}