Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
1 2
1 3

Sample Output 1

2
1
1

The graph in this input is as follows:

For k=1, there are two ways in which we can write the numbers on the vertices, as follows:

Writing 1, 2, 3 on Vertex 1, 2, 3, respectively

Writing 1, 3, 2 on Vertex 1, 2, 3, respectively

Sample Input 2

2
1 2

Sample Output 2

1
1

The graph in this input is as follows:

Sample Input 3

5
1 2
2 3
3 4
3 5

Sample Output 3

2
8
12
3
3

The graph in this input is as follows:

Sample Input 4

8
1 2
2 3
3 4
3 5
3 6
6 7
6 8

Sample Output 4

40
280
840
120
120
504
72
72

The graph in this input is as follows:

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <algorithm>
#include <vector>

const int MAXN = (int)2e5;
const int P = (int)1e9 + 7;

std::vector<int> edges[MAXN];
int size[MAXN];     // size of subtree(i).
int64_t f[MAXN];    // subtree(i), 先选i, 的方案数.
int64_t ans[MAXN];  // 整棵树, 先选i, 的方案数.
int64_t factorial[MAXN];

// inv of P.
int64_t inv(int64_t x) {
  int y = P - 2;
  int64_t ans = 1;
  while (y > 0) {
    if (y & 1) ans = ans * x % P;
    x = x * x % P;
    y >>= 1;
  }
  return ans;
}

// Calc C(x, y).
int64_t comb(int x, int y) {
  if (y == 0 || y == x) return 1;
  return factorial[x] * inv(factorial[y] * factorial[x - y] % P) % P;
}

void calc_f(int u, int pa) {
  size[u] = 0;
  f[u] = 1; 
  for (const auto& v : edges[u]) {
    if (v == pa) continue;
    calc_f(v, u);
    size[u] += size[v];
    f[u] = f[u] * comb(size[u], size[v]) % P * f[v] % P;
  }
  ++size[u];
}

// ans[v] = (size[v] / (n - size[v])) * ans[u].
void calc_ans(int u, int pa, int n) {
  for (const auto& v : edges[u]) {
    if (v == pa) continue;
    ans[v] = ans[u] * size[v] % P * inv(n - size[v]) % P;
    calc_ans(v, u, n);
  }
}

void solve(int n) {
  factorial[0] = 1;
  for (int i = 1; i <= n; ++i) {
    factorial[i] = factorial[i - 1] * i % P;
  }

  calc_f(0, -1);
  ans[0] = f[0];
  calc_ans(0, -1, n);
}

int main() {
  int n;
  std::cin >> n;
  for (int i = 1, u, v; i < n; ++i) {
    std::cin >> u >> v;
    --u, --v;
    edges[u].emplace_back(v);
    edges[v].emplace_back(u);
  }
  solve(n);
  for (int i = 0; i < n; ++i) std::cout << ans[i] << std::endl;

  return 0;
}