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;
typedef double db;
typedef long long ll;
typedef pair<int, int> pii;
#define fi first
#define se second
#define lbd(a, x) lower_bound(all(a), x)
const int mod = 1e9 + 7;
int power_mod(int a, int b) {
  int res = 1;
  for(; b; b >>= 1, a = (ll) a * a % mod)
		if(b & 1) res = (ll) res * a % mod;
	return res;
}
const int N = 2e5 + 10;
vector<int> E[N];
int fac[N], ifac[N];
int dp[N], sz[N], n;
void dfs1(int u, int pre) {
  dp[u] = sz[u] = 1;
  for(auto &v : E[u]) {
		if(v == pre) continue;
		dfs1(v, u);
		sz[u] += sz[v];
    dp[u] = (ll) dp[u] * dp[v] % mod * ifac[sz[v]] % mod;
  }
  dp[u] = (ll) dp[u] * fac[sz[u] - 1] % mod;
}
int up[N];
void dfs2(int u, int pre) {
  for(auto &v : E[u]) {
    if(v == pre) continue;
    up[v] = (ll) dp[u] * power_mod(dp[v], mod - 2) % mod;
    up[v] = (ll) up[v] * ifac[n - 1] % mod * fac[n - sz[v] - 1] % mod * fac[sz[v]] % mod;
    dp[v] = (ll) dp[v] * ifac[sz[v] - 1] % mod * fac[n - 1] % mod;
    dp[v] = (ll) dp[v] * ifac[n - sz[v]] % mod;
    dp[v] = (ll) dp[v] * up[v] % mod;
    dfs2(v, u);
  }
}
int main() {
#ifdef local
	freopen("in.txt", "r", stdin);
#endif
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	fac[0] = ifac[0] = 1;
	for(int i = 1; i < N; i++) fac[i] = (ll) fac[i - 1] * i % mod;
	ifac[N - 1] = power_mod(fac[N - 1], mod - 2);
	for(int i = N - 2; i; i--) ifac[i] = (ll) ifac[i + 1] * (i + 1) % mod;
	cin >> n;
	for(int i = 1, u, v; i < n; i++) {
		cin >> u >> v;
		E[u].push_back(v);
		E[v].push_back(u);
	}
	dfs1(1, 0);
	dfs2(1, 0);
	for(int i = 1; i <= n; i++) {
		cout << dp[i] << '\n';
	}
	return 0;
}
