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 long long ll;
 
const int N = 2e5 + 5, mod = 1e9 + 7;
 
int n;
int siz[N], mul[N], ans[N];
vector<int> g[N];
 
int pw(int x, int y) {
    int res = 1;
    while (y) {
        if (y & 1)
            res = (ll)res * x % mod;
        x = (ll)x * x % mod;
        y >>= 1;
    }
    return res; 
}
 
void dfs(int u, int fa) {
    siz[u] = 1, mul[u] = 1;
    for (auto v : g[u]) 
        if (v != fa) {
            dfs(v, u);
            siz[u] += siz[v];
            mul[u] = (ll)mul[u] * mul[v] % mod;
        }
    mul[u] = (ll)mul[u] * siz[u] % mod;
}
 
void dfs2(int u, int fa, int fromFa) {
    if (u != 1)
        ans[u] = (ll)mul[u] * pw(siz[u], mod - 2) % mod * n % mod * fromFa % mod;
    else 
        ans[u] = (ll)mul[u] * fromFa % mod; 
    for (auto v : g[u]) 
        if (v != fa) {
            int tmp = (ll)fromFa * mul[u] % mod * pw(mul[v], mod - 2) % mod * pw(siz[u], mod - 2) % mod * (n - siz[v]) % mod;
            dfs2(v, u, tmp);
        }
}
 
int main() {
    ios::sync_with_stdio(false);
    cin >> n;
    for (int i = 1; i < n; i++) {
        int u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1, 0);
    dfs2(1, 0, 1);
    int x = 1;
    for (int i = 1; i <= n; i++)
        x = (ll)x * i % mod;
    for (int i = 1; i <= n; i++)
        cout << (ll)x * pw(ans[i], mod - 2) % mod << '\n';
    return 0; 
}