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;

typedef unsigned long ul;
typedef unsigned long long ull;
typedef long long ll;
typedef vector<ll> vint;
typedef vector< vector<ll> > vvint;
typedef vector< vector< vector<ll> > > vvvint;
typedef vector<string> vstring;
typedef vector< vector<string> > vvstring;
typedef vector<char> vchar;
typedef vector< vector<char> > vvchar;
typedef vector<long double> vdouble;
typedef vector< vector<long double> > vvdouble;
typedef vector< vector< vector<long double> > > vvvdouble;
typedef pair<ll,ll> pint;
typedef vector<pint> vpint;
typedef vector<bool> vbool;

#define rep(i,n) for(ll i=0;i<n;i++)
#define repf(i,f,n) for(ll i=f;i<n;i++)
#define repr(i,n) for(ll i=n-1;i>=0;i--)
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define ALL(obj) (obj).begin(), (obj).end()
// #define LLONG_MAX 9223372036854775806
#define vmax(vec) *max_element(vec.begin(), vec.end())
#define vmin(vec) *min_element(vec.begin(), vec.end())
#define vsort(vec) sort(vec.begin(), vec.end())
#define vsortgr(vec) sort(vec.begin(), vec.end(), greater<ll>())
#define MOD 1000000007
// #define MOD 998244353
// #define MOD LLONG_MAX
const double PI=3.14159265358979323846;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

int dy[]={0, 0, 1, -1};
int dx[]={1, -1, 0, 0,};

void printv(vint &v){
    for(auto e:v) cout<<e<<" ";
        cout<<endl;
    }

// 繰り返し二乗法
ll power(ll a, ll b){
    if(a==1)return a;
    // if(a==0)re
    ll res=1;
    while(b>0){
        if(b&1) res=res*a%MOD;
        a=a*a%MOD;
        b>>=1;
    }
    return res;
}

const int MAX=2100000;

ll fact[MAX], fact_inv[MAX];

void init_fact(ll n){
    fact[0]=1;
    // 階乗の計算
    for(ll i=0; i<n; i++) fact[i+1]=fact[i]*(i+1)%MOD;
    fact_inv[n]=power(fact[n], MOD-2);
    // 逆元の計算
    for(ll i=n-1;i>=0;i--) fact_inv[i]=fact_inv[i+1]*(i+1)%MOD;
}

ll comb(ll n, ll r){
    return (fact[n]*fact_inv[r])%MOD*fact_inv[n-r]%MOD;
}

ll perm(ll n, ll r){
    return (fact[n]*fact_inv[n-r])%MOD;
}
template< int mod >
struct ModInt {
  int x;

  ModInt() : x(0) {}

  ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}

  ModInt &operator+=(const ModInt &p) {
    if((x += p.x) >= mod) x -= mod;
    return *this;
  }

  ModInt &operator-=(const ModInt &p) {
    if((x += mod - p.x) >= mod) x -= mod;
    return *this;
  }

  ModInt &operator*=(const ModInt &p) {
    x = (int) (1LL * x * p.x % mod);
    return *this;
  }

  ModInt &operator/=(const ModInt &p) {
    *this *= p.inverse();
    return *this;
  }

  ModInt operator-() const { return ModInt(-x); }

  ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }

  ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }

  ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }

  ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }

  bool operator==(const ModInt &p) const { return x == p.x; }

  bool operator!=(const ModInt &p) const { return x != p.x; }

  ModInt inverse() const {
    int a = x, b = mod, u = 1, v = 0, t;
    while(b > 0) {
      t = a / b;
      swap(a -= t * b, b);
      swap(u -= t * v, v);
    }
    return ModInt(u);
  }

  ModInt pow(int64_t n) const {
    ModInt ret(1), mul(x);
    while(n > 0) {
      if(n & 1) ret *= mul;
      mul *= mul;
      n >>= 1;
    }
    return ret;
  }

  friend ostream &operator<<(ostream &os, const ModInt &p) {
    return os << p.x;
  }

  friend istream &operator>>(istream &is, ModInt &a) {
    int64_t t;
    is >> t;
    a = ModInt< mod >(t);
    return (is);
  }

  static int get_mod() { return mod; }
};

using modint = ModInt< (int)(1e9+7) >;


ll n;
vvint edges;
map<pint, pair<ll, modint>> dp;

void treeDP(ll src, ll to){
    if(dp.count(mp(src, to))) return;
    ll x;
    modint y;
    if(dp.count(mp(0, to))){
        // puts("yay");
        // cout<<src<<" "<<to<<endl;
        // cout<<dp[mp(0,to)].first<<" "<<dp[mp(0,to)].second<<endl;
        tie(x,y) = dp[mp(0, to)];
        y *= fact_inv[n-1];
        y *= fact[n-1 - dp[mp(to, src)].first];
        y *= fact[dp[mp(to, src)].first];
        y /= dp[mp(to, src)].second;
        x = n - dp[mp(to, src)].first;
        // cout<<x<<" "<<y<<endl<<endl;
    }else{
        x = 0;
        y = 1;
        for(auto e:edges[to]){
            if(e == src) continue;
            if(dp.count(mp(to, e)) == 0) treeDP(to, e);
            ll dx;
            modint dy;
            tie(dx, dy) = dp[mp(to,e)];
            x += dx;
            y *= dy;
            y *= fact_inv[dx];
        }
        y *= fact[x];
        x++;
    }
    dp[mp(src, to)] = mp(x, y);
}


int main() {
    cout<<fixed<<setprecision(10);

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

    init_fact(2e5+100);

    queue<ll> q;
    q.push(1);
    set<ll> used;
    while(!q.empty()){
        ll src = q.front(); q.pop();
        if(used.count(src)) continue;
        used.insert(src);
        treeDP(0, src);
        for(auto to:edges[src]){
            if(used.count(to)) continue;
            q.push(to);
        }
    }

    rep(i,n){
        cout<<dp[mp(0, i+1)].second<<endl;
    }


    return 0;

}