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;

const int N = 200005;
const int mod = 1e9+7;
int add(int a, int b) {
	return a + b < mod ? a + b : a + b - mod;
}
int mul(int a, int b) {
	return int(1LL * a * b % mod);
}
int fexp(int b, int e = mod - 2) {
	int ans = 1;
	while(e) {
		if(e & 1) ans = mul(ans, b);
		b = mul(b, b);
		e /= 2;
	}
	return ans;
}

int fat[N], ifat[N];
int C(int n, int k) {
	if(k > n) return 0;
	return mul(mul(fat[n], ifat[n - k]), ifat[k]);
}

int n;
vector<int> g[N];

int dp[N];
int sz[N];


int f(int u, int p) {
	sz[u] = 1;
	for(int v : g[u]) if(v != p) {
		f(v, u);
		sz[u] += sz[v];
	}
	int &ans = dp[u];
	ans = 1;
	int left = sz[u] - 1;
	for(int v : g[u]) if(v != p) {
		ans = mul(ans, mul(C(left, sz[v]), dp[v]));
		left -= sz[v];
	}
	return ans;
}

int ans[N];
void dfs(int u, int p) {
	ans[u] = dp[u];

	for(int v : g[u]) if(v != p) {
		int oldu = dp[u];
		int oldv = dp[v];

		assert(sz[u] == n);
		dp[u] = mul(dp[u], fexp(mul(C(sz[u] - 1, sz[v]), dp[v])));
		sz[u] -= sz[v];
		sz[v] += sz[u];
		dp[v] = mul(dp[v], mul(C(sz[v] - 1, sz[u]), dp[u]));
		assert(sz[v] == n);
		dfs(v, u);
		sz[v] -= sz[u];
		sz[u] += sz[v];
		dp[v] = oldv;
		dp[u] = oldu;
	}
}

int main() {

	fat[0] = 1;
	for(int i = 1; i < N; i++) fat[i] = mul(i, fat[i - 1]);
	ifat[N - 1] = fexp(fat[N - 1]);
	for(int i = N - 2; i >= 0; i--) ifat[i] = mul(i + 1, ifat[i + 1]);

	scanf("%d", &n);

	for(int i = 1; i < n; i++) {
		int a, b;
		scanf("%d %d", &a, &b);
		g[a].push_back(b);
		g[b].push_back(a);
	}

	f(1, 1);

	dfs(1, 1);

	for(int i = 1; i <= n; i++) {
		printf("%d\n", ans[i]);
	}
}
