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;

typedef long long ll;
typedef pair<int, int> ii;

int n;
const int N = 200001;
const int MOD = (int) 1e9 + 7;
ll fact[N];
ll inv[N];
ll dp[N];
int sz[N];
vector<int> v[N];

ll power(ll a, int x) {
  ll res = 1;
  while (x) {
    if (x & 1) res = (res * a) % MOD;
    a = (a * a) % MOD;
    x >>= 1;
  }
  return res;
}

void init() {
  fact[0] = 0;
  fact[1] = 1;
  for (int i = 2; i < N; ++i) {
    fact[i] = fact[i - 1] * i % MOD;
  }

  inv[1] = 1;
  for (int i = 2; i < N; ++i) {
    inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;
  }
}

void dfs(int x, int par) {
  sz[x] = 1;
  for (auto y : v[x]) {
    if (y == par) continue;
    dfs(y, x);
    sz[x] += sz[y];
  }
}

void reroot(int x, int par) {
  dp[x] = dp[par] * sz[x] % MOD * inv[n - sz[x]] % MOD;
  for (auto y : v[x]) {
    if (y == par) continue;
    reroot(y, x);
  }
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  init();

  int x, y;
  cin >> n;
  for (int i = 1; i < n; ++i) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }

  dfs(1, -1);
  dp[1] = fact[n];
  for (int i = 1; i <= n; ++i) {
    dp[1] = (dp[1] * inv[sz[i]]) % MOD;
  }
  for (auto p : v[1]) {
    reroot(p, 1);
  }
  for (int i = 1; i <= n; ++i) {
    cout << dp[i] << endl;
  }

  return 0;
}
