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 long long ll;
typedef pair<ll, ll> P;
struct edge{ll to,cost;};
const int inf = 1<<27;
const ll INF = 1LL<<60;
const int COMBMAX = 1001001;
const ll MOD = 1000000007;
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define eachdo(v, e) for(const auto& e : (v))
#define all(v) (v).begin(), (v).end()
#define lower_index(v, e) (ll)distance((v).begin(), lower_bound((v).begin(), (v).end(), e))
#define upper_index(v, e) (ll)distance((v).begin(), upper_bound((v).begin(), (v).end(), e))
ll mpow(ll a, ll n, ll mod = MOD){ll res = 1; while(n > 0){if(n & 1)res = res * a % mod; a = a * a % mod; n >>= 1;} return res;}
void YesNo(bool j){cout << (j ? "Yes" : "No") << endl; return;}
void yesno(bool j){cout << (j ? "yes" : "no") << endl; return;}
template<class Head> void pt(Head&& head){cout << head << endl; return;}
template<class Head, class... Tail> void pt(Head&& head, Tail&&... tail){cout << head << " "; pt(forward<Tail>(tail)...);}
template<class T> void debug(T v){rep(i, v.size()) cout << v[i] << " " ; cout << endl;}
template<class T> void debug2(T v){rep(i, v.size()){rep(j, v[i].size()) cout << v[i][j] << " " ; cout << endl;}}
template<class T1, class T2> ll bcount(T1 v, T2 a){return upper_index(v, a) - lower_index(v, a);} 
template<class T1, class T2> inline bool chmin(T1 &a, T2 b){if(a > b){a = b; return true;} return false;}
template<class T1, class T2> inline bool chmax(T1 &a, T2 b){if(a < b){a = b; return true;} return false;}

struct mint {
ll x;
    mint(ll x=0):x((x%MOD+MOD)%MOD){}
    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 {mint res(*this);return res+=a;}
    mint operator-(const mint a) const {mint res(*this);return res-=a;}
    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;}
    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;}
};

class combination {
    public:
    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;
        if (COMBMAX < n){
            mint ret = 1;
            for(ll i = 1; i <= k; i++){
                ret *= (n-i+1);
                ret /= i;
            }
            return ret;
        }
        return fact[n]*ifact[k]*ifact[n-k];
    }

    mint multi(ll n,ll sum,ll l,ll r){
        ll m = r - l + 1;
        ll t = sum - n*(l-1);
        mint ans = 0;
        for(ll k = 0; k <= (t - n) / m; k++){
            ll sign = k%2==0 ? 1 : -1;
            mint temp = this->operator()(n, k);
            temp *= this->operator()(n+(t-n-m*k-1),t-n-m*k);
            temp *= sign;
            ans += temp;
        }
        return ans;
    }
} com(COMBMAX);

ll N;
vector<vector<ll>> g(200005, vector<ll>());
vector<pair<mint, ll>> DPpos(200005);
vector<pair<mint, ll>> DPngt(200005);
vector<bool> posVisited(200005, false);
vector<bool> negVisited(200005, false);

pair<mint, ll> dfs1(ll v, ll par = -1){
    if (posVisited[v]) return DPpos[v];
    mint ret = 1;
    ll siz = 0;
    eachdo(g[v], u){
        if (u == par) continue;
        ret *= com(siz + dfs1(u, v).second, siz); 
        ret *= dfs1(u, v).first;
        siz += dfs1(u, v).second;
    }
    siz += 1;
    DPpos[v] = make_pair(ret, siz);
    posVisited[v] = true;
    return DPpos[v];
}

void dfs2(ll v, ll par = -1){
    // pt(v, par);
    if (par == -1){
        DPngt[v] = DPpos[v];
    }else{
        mint part;
        ll partSize;
        tie(part, partSize) = DPngt[par];

        part /= com(N - 1, DPpos[v].second);
        part /= DPpos[v].first;

        partSize -= DPpos[v].second;

        mint ret;
        ll size;
        tie(ret, size) = DPpos[v];
        size--;
        ret *= com(N - 1, size);
        ret *= part;
        DPngt[v] =  make_pair(ret, N);
    }
    eachdo(g[v], u){
        if (par == u) continue;
        dfs2(u, v);
    }
    return;
}

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    cin >> N;
    rep(i, N - 1){
        ll u, v; cin >> u >> v;
        g[--u].emplace_back(--v);
        g[v].emplace_back(u);
    }
    dfs1(0);
    dfs2(0);
    // rep(i, N) pt(DPpos[i].first.x, DPpos[i].second);
    rep(i, N) pt(DPngt[i].first.x); 
    return 0;
}