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;

int n, sz[200005], mod = 1e9+7;
vector <int> g[200005];
long long ans[200005], inv[200005];

void build(int x){
    inv[1] = inv[0] = 1;
    for(int i = 2 ; i <= x; i++){
        inv[i] = (mod - mod / i * inv[mod % i] % mod) % mod;
    }
}
int dfs(int x, int pre){
    sz[x] = 1;
    for (int i:g[x]){
        if (i == pre) continue;
        sz[x] += dfs(i, x);
    }
    return sz[x];
}
void dfs2(int x, int pre){
    for (int i:g[x]){
        if (i == pre) continue;
        ans[i] = ans[x]*sz[i]%mod*inv[n-sz[i]]%mod;
        dfs2(i, x);
    }
}

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> n;
    int x, y;
    for (int i = 1; i < n; i++){
        cin >> x >> y;
        g[x].push_back(y);
        g[y].push_back(x);
    }
    dfs(1, 0);
    build(n);
    ans[1] = 1;
    for (int i = 1; i <= n; i++){
        ans[1] *= i;
        ans[1] %= mod;
        ans[1] *= inv[sz[i]];
        ans[1] %= mod;
    }
    dfs2(1, 0);
    for (int i = 1; i <= n; i++){
        cout << ans[i] << "\n";
    }
}
