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 pb push_back
#define all(a) (a).begin(), (a).end()

using ll = long long;
using pii = pair<int, int>;

const int N = 2e5 + 10;
const ll MOD = 1e9 + 7;

vector<int> g[N];
int n;
ll ans[N], m[N];
ll total;

ll binpow(ll a, ll b, const ll m = MOD) {
    a %= m;
    ll res = 1;
    while (b > 0) {
        if (b & 1)
            res = res * a % m;
        a = a * a % m;
        b >>= 1;
    }
    return res;
}

ll mod_inverse(int x, const ll m = MOD) {
	return binpow(x, m - 2, m);
}

ll mul(ll a, ll b, const ll m = MOD) {
	return a * b % m;
}

int dfs1(int u, int p = -1) {
	int sz = 1;
	for (auto v : g[u]) {
		if (v != p) {
			sz += dfs1(v, u);  
		}
	}
	if (p != -1) {
		total = mul(total, mod_inverse(sz));
		m[u] = mul(m[u], sz);
		m[u] = mul(m[u], mod_inverse(n - sz));
	}
	return sz;
}

void dfs2(int u, int p = -1, ll val = 1) {
	val = mul(val, m[u]);
	ans[u] = mul(total, val);
	for (auto v : g[u]) {
		if (v != p)
			dfs2(v, u, val);
	}
}

int main() {
    ios::sync_with_stdio(false);cin.tie(nullptr);
    cin >> n;
    for (int i = 0; i < n - 1; ++i) {
    	int a, b; cin >> a >> b;
    	g[a].pb(b);
    	g[b].pb(a);
    }

    fill_n(m + 1, n, 1LL);
    total = 1;
    for (int i = 2; i < n; ++i)
    	total = mul(total, i);
    dfs1(1);
    dfs2(1);

    for (int i = 1; i <= n; ++i) {
    	cout << ans[i] << '\n';
    }
}