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;
#define ll long long
const int N = 2e5 + 10;
const int INF = 0x3f3f3f3f;
const ll mod = 1e9 + 7;
int n;
ll fac[N], inv[N];
ll power(ll a, ll x) {
	ll ans = 1;
	while (x) {
		if (x & 1) ans = (ans * a) % mod;
		a = (a * a) % mod;
		x >>= 1;
	}
	return ans;
}
void init() {
	fac[0] = 1;
	for (int i = 1; i < N; i++) {
		fac[i] = fac[i - 1] * i %mod;
	}
	inv[N - 1] = power(fac[N - 1], mod - 2);
	for (int i = N - 2; i >= 0; i--) {
		inv[i] = inv[i + 1] * (i + 1) % mod;
	}
}
ll C(ll n, ll k) {
	return fac[n] * inv[k] % mod*inv[n - k] % mod;
}
vector<int>G[N];
ll dp[N], ans[N], cnt[N];
void dfs1(int u, int _fa) {
	cnt[u] = dp[u] = 1;
	for (int v : G[u]) {
		if (v == _fa)continue;
		dfs1(v, u);
		cnt[u] += cnt[v];
		dp[u] = dp[u] * dp[v] % mod * C(cnt[u] - 1, cnt[v]) % mod;
	}
}
void dfs2(int u, int _fa) {
	for (int v : G[u]) {
		if (v == _fa)continue;
		ans[v] = ans[u] * power(dp[v] * C(n - 1, cnt[v])%mod, mod - 2) % mod;
		ans[v] = ans[v] * dp[v] % mod*C(n - 1, n - cnt[v])%mod;
		dfs2(v, u);
	}
}
int main() {
	init();
	scanf("%d", &n);
	for (int i = 1; i < n; i++) {
		int u, v;
		scanf("%d%d", &u, &v);
		G[u].push_back(v);
		G[v].push_back(u);
	}
	dfs1(1, 0);
	ans[1] = dp[1];
	dfs2(1, 0);
	for (int i = 1; i <= n; i++)printf("%lld\n", ans[i]);
	return 0;
}