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;

const int MaxN = 200005;
const long long Mod = 1000000007;

int n;
vector<int> Adj[MaxN];
long long gt[MaxN], f[MaxN], sz[MaxN];

long long Power(long long a, long long b)
{
    if (b == 0)
    {
        return 1;
    }
    long long t = Power(a, b / 2);
    t = (t * t) % Mod;
    if (b % 2 == 1)
    {
        t = (t * a) % Mod;
    }
    return t;
}

void DFS1(int u, int p)
{
    sz[u] = 1;
    for (auto v : Adj[u])
    {
        if (v != p)
        {
            DFS1(v, u);
            sz[u] += sz[v];
        }
    }
}
void DFS2(int u, int p)
{
    for (auto v : Adj[u])
    {
        if (v != p)
        {
            f[v] = (f[u] * sz[v]) % Mod;
            sz[u] = n - sz[v];
            sz[v] = n;
            f[v] = (f[v] * Power(sz[u], Mod - 2)) % Mod;
            DFS2(v, u);
            sz[v] = n - sz[u];
            sz[u] = n;
        }
    }
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n;
    int u, v;
    for (int i = 1; i < n; i++)
    {
        cin >> u >> v;
        Adj[u].push_back(v);
        Adj[v].push_back(u);
    }
    gt[0] = 1;
    for (int i = 1; i <= n; i++)
    {
        gt[i] = (gt[i - 1] * i) % Mod;
    }
    DFS1(1, -1);
    f[1] = gt[n];
    for (int i = 1; i <= n; i++)
    {
        f[1] = (f[1] * Power(sz[i], Mod - 2)) % Mod;
    }
    DFS2(1, -1);
    for (int i = 1; i <= n; i++)
    {
        cout << f[i] << '\n';
    }
    return 0;
}
