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>

typedef long long ll;

const ll kMod = 1e9 + 7;
const int kN = 2e5 + 5;

ll QPow(ll a, ll b) {
	ll ans = 1, bas = a % kMod;
	while(b) {
		if(b & 1) ans = ans * bas % kMod;
		bas = bas * bas % kMod;
		b >>= 1;
	}
	return ans;
}

std::vector<int> E[kN];
int n, siz[kN], fa[kN]; ll f[kN];
void Dfs1(int x) {
	f[x] = 1; siz[x] = 1;
	for(int i : E[x])
		if(!siz[i]) {
			Dfs1(i); siz[x] += siz[i]; fa[i] = x;
			f[x] = f[x] * f[i] % kMod;
		}
	f[x] = f[x] * siz[x] % kMod;
}
void Dfs2(int x) {
	for(int i : E[x])
		if(i != fa[x]) {
			f[i] = f[x] * (n - siz[i]) % kMod * QPow(siz[i], kMod - 2) % kMod;
			Dfs2(i);
		}
}

int main() {
	scanf("%d", &n);
	ll fac = 1;
	for(int i = 1; i <= n; ++i) fac = fac * i % kMod;
	for(int i = 1; i < n; ++i) {
		int a, b; scanf("%d%d", &a, &b);
		E[a].push_back(b); E[b].push_back(a);
	}
	Dfs1(1); Dfs2(1);
	for(int i = 1; i <= n; ++i)
		printf("%lld\n", fac * QPow(f[i], kMod - 2) % kMod);
	return 0;
}