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 MAX_N = 200010;
long long mod = 1e9 + 7;

vector<int> G[MAX_N];
vector<int> sz(MAX_N);
vector<long long> ans(MAX_N);
long long res1 = 1;
int n;

long long inv(long long a, long long p){
    long long res = 1;
    while (p){
        if (p & 1LL)
            res = (res * a) % mod;
        a = ((a % mod) * (a % mod)) % mod;
        p >>= 1;
    }

    return res;
}

void dfs(int u, int p){
    sz[u] = 1;
    for (auto v: G[u]){
        if (v != p){
            dfs(v, u);
            sz[u] += sz[v];
        }
    }

    res1 = ((res1 % mod) * (sz[u] % mod)) % mod;
}

void dfs1(int u, int p){
    if (p != -1){
        ans[u] = (((ans[p] % mod) * (sz[u] % mod) % mod) * (inv(n - sz[u], mod - 2) % mod)) % mod;
    }

    for (auto v: G[u]){
        if (v != p){
            dfs1(v, u);
        }
    }
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    cin >> n;
    
    for (int i = 0; i < n - 1; i++){
        int u, v;
        cin >> u >> v;
        u--; v--;

        G[u].push_back(v);
        G[v].push_back(u);
    }

    dfs(0, -1);
    ans[0] = 1;
    for (long long i = 1; i <= n; i++)
        ans[0] = ((ans[0] % mod) * (i % mod)) % mod;
    ans[0] = (ans[0] * inv(res1, mod - 2)) % mod;

    dfs1(0, -1);

    for (int i = 0; i < n; i++)
        cout << ans[i] << '\n';

    return 0;
}
