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 int long long
typedef pair<int,int> pi;
#define f first
#define s second
#define FAST ios_base::sync_with_stdio(0); cin.tie(0);
#define all(x) x.begin(),x.end()
const int maxn = 200010;
const int mod = 1e9 + 7;

int n;
vector <int> adjlist[maxn];
int fac[maxn];
map <int,int> mp;
int dp[maxn],sz[maxn],ans[maxn];

int expo(int x, int exp) {
	int res = 1, base = x;
	while (exp) {
		if (exp & 1) res = (res * base) % mod;
		base = (base*base) % mod;
		exp >>= 1;
	}
	return res;
}

int inv(int x) {
	if (mp.find(x) != mp.end()) return mp[x];
	else return mp[x] = expo(x,mod-2);
}


int choose(int N, int k) {
	return fac[N] * inv(fac[k]*fac[N - k] % mod) % mod;
}

void dfsf(int x, int p) {
	sz[x] = dp[x] = 1;
	for (auto i : adjlist[x]) {
		if (i == p) continue;
		dfsf(i,x);
		sz[x] += sz[i];
		dp[x] *= dp[i] * choose(sz[x]-1,sz[i]) % mod;
		dp[x] %= mod;
		//cout << sz[x] << " " << sz[i] << "\n";
		//cout << choose(sz[x] -1, sz[i]) << "\n";
		//cout << i << " " << dp[x] << "\n";
	}
}

void reroot(int x, int p) {
	for (auto i: adjlist[x]) {
		if (i == p) continue;
		ans[i] = ans[x] * inv(dp[i] * choose(n-1,sz[i]) % mod) % mod * choose(n-1,n - sz[i]) % mod;
		ans[i] *= dp[i];
		ans[i] %= mod;
		reroot(i,x);
	}
}

int32_t main() {
	//FAST
	cin >> n;
	for (int i =0;i<n-1;i++) {
		int a,b; cin >> a >> b;
		adjlist[a].push_back(b);
		adjlist[b].push_back(a);
	}
	
	fac[0] = 1;
	for (int i =1;i<maxn;i++) {
		fac[i] = fac[i-1] * i % mod;
	}
	
	dfsf(1,-1);
	ans[1] = dp[1];
	reroot(1,-1);
	
	for (int i =1;i<=n;i++) cout << ans[i] << "\n";
	
}







