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 LL long long
using namespace std;
const int N = 2e5 + 7;
const LL MOD = 1e9 + 7;

LL f[N], g[N], fac[N], inv[N];
vector<int > p[N];
int sz[N], n;

LL qpow(LL x, LL y){
	LL z = 1;
	while (y){
		if (y & 1) z = z * x % MOD;
		x = x * x % MOD;
		y /= 2;
	}
	return z;
}

void gao1(int x, int fa){
	sz[x] = 1; f[x] = 1;
	for (int i = 0; i < p[x].size(); i++) if (p[x][i] != fa){
		gao1(p[x][i], x);
		sz[x] += sz[p[x][i]];
		f[x] = f[x] * f[p[x][i]] % MOD;
		f[x] = f[x] * inv[sz[p[x][i]]] % MOD;
	}
	f[x] = f[x] * fac[sz[x] - 1] % MOD;
}

void gao2(int x, int fa){
	for (int i = 0; i < p[x].size(); i++) if (p[x][i] != fa){
		LL t = f[x] * qpow(f[p[x][i]], MOD - 2) % MOD * fac[sz[p[x][i]]] % MOD;
		t = t * inv[sz[x] - 1] % MOD;
		t = t * fac[n - sz[p[x][i]] - 1] % MOD;
		if (x != 1) t = t * g[x] % MOD * inv[n - sz[x]] % MOD;
		g[p[x][i]] = t;
		gao2(p[x][i], x);
	}
}

int main(){
	scanf("%d", &n);
	fac[0] = 1;
	for (int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % MOD;
	for (int i = 0; i <= n; i++) inv[i] = qpow(fac[i], MOD - 2);

	for (int i = 1; i < n; i++){
		int u, v; scanf("%d%d", &u, &v);
		p[u].push_back(v); p[v].push_back(u);
	}
	gao1(1, 0);
	gao2(1, 0);
	for (int i = 1; i <= n; i++){
		if (i == 1) printf("%lld\n", f[1]);
		else {
			LL t = f[i] * g[i] % MOD * inv[n - sz[i]] % MOD;
			t = t * inv[sz[i] - 1] % MOD * fac[n - 1] % MOD;
			printf("%lld\n", t); 
		}
	}	
	return 0;
}