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 mod = 1e9 + 7;

int pw(int x, int y) {
    int r = 1;
    while (y) {
        if (y & 1) r = 1ll * r * x % mod;
        x = 1ll * x * x % mod;
        y >>= 1;
    }
    return r;
}
int inv(int x) {
    return pw(x, mod - 2);
}

int n;
vector <int> adj[200005];
int dad[200005];
int sz[200005];
int ans[200005];
int add = 1;

void dfs(int u) {
    sz[u] = 1;
    for (auto v: adj[u]) {
        if (v == dad[u]) continue;
        dad[v] = u;
        dfs(v);
        sz[u] += sz[v];
    }
    add = 1ll * add * inv(sz[u]) % mod;
}
void dfs2(int u) {
    ans[u] = add;
    for (auto v: adj[u]) {
        if (v == dad[u]) continue;
        int save = add;
        add = 1ll * add * sz[v] % mod;
        add = 1ll * add * inv(n - sz[v]) % mod;
        dfs2(v);
        add = save;
    }
}

int main(void) {
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    cin >> n;
    for (int i = 1; i < n; ++i) {
        int u, v;
        cin >> u >> v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    for (int i = 1; i <= n; ++i) add = 1ll * add * i % mod;
    dfs(1);
    dfs2(1);
    for(int i=1;i<=n;i++)cout<<ans[i]<<"\n";
    return 0;
}