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 p = 1e9 + 7;
const int N = 2e5 + 5;
int n, x, y, frac[N], inv[N], ans, dp[N], cnt, head[N], sz[N];
struct node { int u, v; } edge[N << 1]; 
inline void add(int x, int y) { edge[cnt].u = head[x], edge[cnt].v = y, head[x] = cnt++; }
inline int power(int x, int y, int ret = 1) { for (; y; y >>= 1, x = (ll) x * x % p) if (y & 1) ret = (ll) ret * x % p; return ret; }
void dfs(int x, int fa) {
	int i;
	sz[x] = 1;
	for (i = head[x]; ~i; i = edge[i].u) {
		int y = edge[i].v;
		if (y == fa) continue;
		dfs(y, x);
		sz[x] += sz[y], dp[x] = (ll) dp[x] * dp[y] % p * inv[sz[y]] % p;
	}
	dp[x] = (ll) dp[x] * frac[sz[x] - 1] % p;
}
void solve(int x, int fa) {
	int i;
	if (fa) dp[x] = (ll) dp[fa] % p * sz[x] % p * inv[n - sz[x]] % p * frac[n - sz[x] - 1] % p;
	for (i = head[x]; ~i; i = edge[i].u) {
		int y = edge[i].v;
		if (y == fa) continue;
		solve(y, x);
	}
}
int main() {
	memset(head, -1, sizeof head);
	int i;
	std::cin >> n;
	frac[0] = frac[1] = dp[1] = 1;
	for (i = 2; i <= n; ++i) frac[i] = (ll) frac[i - 1] * i % p, dp[i] = 1;
	inv[n] = power(frac[n], p - 2);
	for (i = n; i; --i) inv[i - 1] = (ll) i * inv[i] % p;
	for (i = 1; i < n; ++i) std::cin >> x >> y, add(x, y), add(y, x);
	dfs(1, 0), solve(1, 0);
	for (i = 1; i <= n; ++i) std::cout << dp[i] << '\n';
}