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: Time Limit Exceeded)

#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <unordered_map>
#include <vector>
#include <string.h>
#include <set>

using namespace std;

#define COUT(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << endl

template<class T> void printvec(const vector<T>& v) {
  for (auto x : v) { cout << x << " "; }
  cout << endl;
}
template<class T> void printtree(const vector< vector<T> >& tree) {
  for (long long i = 0; i < tree.size(); ++i) {
    cout << i + 1 << ": "; printvec(tree[i]);
  }
}
template<class T, class U> void printmap(const map<T, U>& mp) {
  for (auto x : mp) { cout << x.first << "=>" << x.second << endl; }
}

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

#define rep(i, n) for(ll i = 0; i < n; ++i)
#define all(s) s.begin(), s.end()
#define fr first
#define sc second
#define mp make_pair
#define pb push_back

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;
typedef tuple<ll, ll, ll> triple;
typedef double D;

const ll INF = 1e9;
const ll MOD = 1000000007;  // 1e9 + 7

// Mod int
// cf. https://www.youtube.com/watch?v=1Z6ofKN03_Y
struct mint {
  ll x;
  mint(ll x = 0) : x((x + MOD) % MOD) {}
  mint& operator+= (const mint a) {
    if ((x += a.x) >= MOD) x %= MOD;
    return *this;
  }
  mint operator+ (const mint a) const {
    mint res(*this);
    return res += a;
  }
  mint& operator-= (const mint a) {
    if ((x += MOD - a.x) >= MOD) x %= MOD;
    return *this;
  }
  mint operator- (const mint a) const {
    mint res(*this);
    return res -= a;
  }
  mint& operator*= (const mint a) {
    (x *= a.x) %= MOD;
    return *this;
  }
  mint operator* (const mint a) const {
    mint res(*this);
    return res *= a;
  }
  mint pow(ll t) const {
    if (!t) { return 1; }
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const {
    return pow(MOD-2);
  }
  mint& operator/= (const mint a) {
    return (*this) *= a.inv();
  }
  mint operator/ (const mint a) const {
    mint res(*this);
    return res /= a;
  }
};

// int main(int argc, char** argv) {
//   // int p;
//   // cin >> p;
//
//   MOD = 13;
//   mint p(10);
//   cout << (p + 15).x << endl;   // 12 (25 % 13)
//   cout << (p - 15).x << endl;   // 8  (-5 % 13)
//   cout << (p * 2).x << endl;    // 7  (20 % 13)
//   cout << (p.pow(3)).x << endl; // 12 (1000 % 13)
//   cout << (p / 3).x << endl;    // 12 (12 * 3 = 10 (36 % 13))
//
//   mint p2(-3);
//   cout << p2.x << endl; // 10 (-3 % 13)
// }

// Combination mod prime.
// cf. https://www.youtube.com/watch?v=1Z6ofKN03_Y
struct Combination {
  vector<mint> fact, ifact;
  Combination(int n) { init(n); }
  void init(int n) {
    assert(n < MOD); // n must be lower than MOD.

    fact.resize(n + 1);
    ifact.resize(n + 1);

    fact[0] = 1;
    for (int i = 1; i <= n; ++i) { fact[i] = fact[i-1] * i; }
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i) { ifact[i-1] = ifact[i] * i; }
  }
  mint operator() (int n, int k) const {
    if (k < 0 || k > n) { return 0; }
    return fact[n] * ifact[k] * ifact[n-k];
  }
  mint perm(int n, int k) const {
    if (k < 0 || k > n) { return 0; }
    return fact[n] * ifact[n-k];
  }
};

Combination c(1);

// int main(int argc, char** argv) {
//   MOD = 13;
//
//   Combination c(12);
//   cout << c(12, 0).x << endl;  // 1  = 1 % 13
//   cout << c(12, 1).x << endl;  // 12 = 12 % 13
//   cout << c(12, 2).x << endl;  // 1  = 66 % 13 = (12 * 11 / 2) % 13
//   cout << c(12, 3).x << endl;  // 12 = 220 % 13 = (12 * 11 * 10 / (3 * 2 * 1)) % 13
//   cout << c(12, 11).x << endl; // 12 = 12 % 13
//   cout << c(12, 12).x << endl; // 1  = 12 % 13
//
//   cout << c.perm(12, 3).x << endl; // 7 = 12 * 11 * 10 % 13
// }

vector<vector<ll>> g; // tree

vector<mint> ans;

// vector<vector<pair<ll, mint>>> subCnts; // cnts[i][j] .. parent i to subtree j
// vector<vector<pair<ll, mint>>> subCnts; // cnts[i][j] .. parent i to subtree j

map<ll, map<ll, pair<ll, mint>>> subCnts;

// return the count of subtree
pair<ll, mint> dfs(int v, int p) {
  if (p != -1 && subCnts[p].find(v) != subCnts[p].end()) {
    return subCnts[p][v];
  }

  vector<ll> cnts;
  vector<mint> patterns;
  vector<ll> subs;

  for (int nv : g[v]) {
    if (nv == p) { continue; } // skip parent
    auto p = dfs(nv, v); // { cnt, pattern }
    cnts.push_back(p.first);
    patterns.push_back(p.second);
    subs.push_back(nv);
  }
  // cout << "v:" << v << endl;
  // cout << "cnts:";
  // printvec(cnts);
  // cout << "patterns:";
  // rep(i, patterns.size()) {
  //   cout << patterns[i].x << " ";
  // }
  // cout << endl;

  ll totalCnt = 0;
  mint totalPattern(1);
  rep(i, cnts.size()) {
    ll cnt = cnts[i];
    totalCnt += cnt;
  }
  totalPattern *= c.fact[totalCnt];
  // cout << "perm(2): " << c.fact[2].x << endl;

  rep(i, cnts.size()) {
    ll cnt = cnts[i];
    mint p = patterns[i];

    totalPattern *= p;
    totalPattern /= c.fact[cnt];
  }

  // Memorize at root
  if (p == -1) {
    // Here, we can calculate subCnts[subV][v] easily
    rep(i, cnts.size()) {
      int nv = subs[i];
      int cnt = cnts[i];
      mint p = patterns[i];

      mint subP = totalPattern;
      subP *= c.fact[cnt];
      subP /= c.fact[totalCnt];
      subP *= c.fact[totalCnt - cnt];
      subP /= p;

      subCnts[nv][v] = { totalCnt + 1 - cnt, subP };
    }
  }

  // cout << "totalCnt: " << totalCnt << endl;
  // cout << "totaPattern: " << totalPattern.x << endl;

  subCnts[p][v] = { totalCnt + 1, totalPattern };
  return subCnts[p][v];
}

int main(int argc, char** argv) {
  cin.tie(NULL);
  cout.tie(NULL);
  ios_base::sync_with_stdio(false);
  //cout << setprecision(10) << fixed;

  ll n;
  cin >> n;
  ans.resize(n);

  // subCnts = vector<vector<pair<ll, mint>>(n+1, vector<pair<ll, mint>>(n+1, pair<ll, mint>(INF, mint(0)));

  c.init(n+5);

  g.resize(n);
  rep(i, n-1) {
    ll a, b;
    cin >> a >> b;
    --a; --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }

  // Here, we should do tree dp.
  rep(i, n) {
    auto p = dfs(i, -1);
    // Here, p is the result from 0
    ll cnt = p.first;
    mint ans = p.second;
    cout << ans.x << endl;
  }
}