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 <bits/stdc++.h>

using namespace std;

using ll = long long;
using pll = pair<ll, ll>;
using vl = vector<ll>;
using vll = vector<vl>;
using vpll = vector<pll>;

void debug_out() { cout << endl; }
 
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cout << H << " ";
  debug_out(T...);
}

#ifdef LOCAL
#define debug(...) cout << "debug: "; debug_out(__VA_ARGS__)
#else
#define debug(...) 
#endif

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rrep(i, n) for (int i = ((int)(n)-1); i >= 0; i--)
#define Rep(i, n) for (long long i = 0; i< (long long)(n); i++)
#define RRep(i,n) for (long long i = ((long long)(n)-1); i>=0; i--)
#define all(v) (v).begin(),(v).end()

struct Edge{
    int to;
    int weight;
    Edge(int t, int w) : to(t), weight(w){ }
};

struct edge{
    int from;
    int to;
    int weight;
    edge(int f,int t,int w) : from(f), to(t), weight(w){ }
};


using Graph = vector<vector<Edge>>;
using graph = vector<vector<int>>;
using edges = vector<edge>;

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

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

template <class T>
T gcd(T a, T b){
    if( b==0 ) return a;
    return gcd(b, a%b);
}

template <class T>
T lcm(T a, T b){
    return (a*b)/gcd(a,b);
}

bool isPalindrome(string s){
    string t = s;
    reverse(all(t));
    return s==t;
}

constexpr int INF = 1e9+7;

vector<int> dx = {-1,0,1,0};
vector<int> dy = {0,-1,0,1};

const int mod = 1000000007;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  struct combination {
  vector<mint> fact, ifact;
  combination(int n):fact(n+1),ifact(n+1) {
    assert(n < mod);
    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) {
    if (k < 0 || k > n) return 0;
    return fact[n]*ifact[k]*ifact[n-k];
  }
};

  // 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 { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}

struct combination {
  vector<mint> fact, ifact;
  combination(int n):fact(n+1),ifact(n+1) {
    assert(n < mod);
    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) {
    if (k < 0 || k > n) return 0;
    return fact[n]*ifact[k]*ifact[n-k];
  }
} comb(200005);

struct Rerooting {
  struct DP {
    mint dp;
    int t;
    DP(mint dp=1, int t = 0): dp(dp),t(t) {}
    DP operator+(const DP& a) const {
      return DP(dp*a.dp*comb(t+a.t,t), t+a.t);
    }
    DP addRoot() const {
      return DP(dp,t+1);
    }
  };
  
  int n;
  vector<vector<int>> to;
  vector<vector<DP>> dp;
  vector<DP> ans;
  Rerooting(int n=0):n(n),to(n),dp(n),ans(n) {}
  void addEdge(int a, int b) {
    to[a].push_back(b);
    to[b].push_back(a);
  }
  void init() {
    dfs(0);
    bfs(0);
  }

  DP dfs(int v, int p=-1) {
    DP dpSum;
    dp[v] = vector<DP>(to[v].size());
    rep(i,to[v].size()) {
      int u = to[v][i];
      if (u == p) continue;
      dp[v][i] = dfs(u,v);
      dpSum = dpSum + dp[v][i];
    }
    return dpSum.addRoot();
  }
  void bfs(int v, const DP& dpP=DP(), int p=-1) {
    int deg = to[v].size();
    rep(i,deg) if (to[v][i] == p) dp[v][i] = dpP;

    vector<DP> dpSumL(deg+1);
    rep(i,deg) dpSumL[i+1] = dpSumL[i] + dp[v][i];
    vector<DP> dpSumR(deg+1);
    for (int i = deg-1; i >= 0; --i)
      dpSumR[i] = dpSumR[i+1] + dp[v][i];
    ans[v] = dpSumL[deg].addRoot();

    rep(i,deg) {
      int u = to[v][i];
      if (u == p) continue;
      DP d = dpSumL[i] + dpSumR[i+1];
      bfs(u, d.addRoot(), v);
    }
  }
};

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    Rerooting r(n);
    rep(i,n-1){
        int a,b;
        cin >> a >> b;
        a--;
        b--;
        r.addEdge(a,b);
    }
    r.init();
    rep(i,n){
        cout << r.ans[i].dp << endl;
    }
    return 0;
}