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++ (GCC 9.2.1) (judge verdict: Accepted)


#pragma GCC optimize ("O3")

#include <bits/stdc++.h>

using namespace std;

#ifdef LOCAL
#include <prettyprint.hpp>
#define debug(...)  cerr << "[" << #__VA_ARGS__ << "]: ", d_err(__VA_ARGS__);
#else
#define debug(...) 83;
#endif

void d_err() {
    cerr << endl;
}

template <typename H, typename... T>
void d_err(H h, T... t) {
    cerr << h << " ";
    d_err(t...);
}

template <typename T>
void print(T x) {
    cout << x << endl;
}

#define ALL(x) (x).begin(), (x).end()
#define FOR(i, m, n) for (int i = (m); i < (n); ++i)
#define REP(i, n) FOR(i, 0, n)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define bcnt __builtin_popcountll

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef pair<ll,ll> Pll;
typedef pair<int,int> Pin;

ll INF = 1e16;
int inf = 1e9;
ll MOD = 1e9+7;

const int MAX_N = 200005;

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 {
        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;
    }
    // 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;
    }
};

ostream& operator << (ostream& os, const mint a) {
    os << a.x;

    return os;
}

struct Comb {
    vector<ll> fact, finv, inv;

    Comb(int nmax): fact(nmax + 1), finv(nmax + 1), inv(nmax + 1) {
        REP(i, 2) {
            fact[i] = 1;
            finv[i] = 1;
            inv[i] = 1;
        }

        FOR(i, 2, nmax + 1) {
            fact[i] = fact[i-1] * i % MOD;
            inv[i] = MOD - MOD / i * inv[MOD % i] % MOD;
            finv[i] = finv[i-1] * inv[i] % MOD;
        }
    }

    mint operator()(int n, int k) {
        if (n < k || n < 0 || k < 0) return 0;

        return mint(fact[n]) * finv[k] * finv[n-k];
    }

    mint perm(int n, int k) {
        if (n < k || n < 0 || k < 0) return 0;

        return mint(fact[n]) * finv[n-k];
    }

} comb(MAX_N);

vector<vi> to(MAX_N);

struct DP {
    mint dp;
    int t;
    DP(mint dp=1, int t=0): dp(dp), t(t) {}
    DP& operator+=(const DP& a) {
        dp *= a.dp;
        dp *= comb(t + a.t, t);
        t += a.t;
        return *this;
    }
    DP operator-(const DP& a) const {
        DP res(*this);
        res.t -= a.t;
        res.dp /= comb(res.t+a.t, res.t);
        res.dp /= a.dp;
        return res;
    }
    DP addRoot() const {
        DP res(*this);
        res.t++;
        return res;
    }
};

vector<DP> dp(MAX_N);

void dfs(int v, int p=-1) {
    for(int u : to[v]) {
        if (u == p) continue;
        dfs(u, v);
        dp[v] += dp[u].addRoot();
    }
}

void bfs(int v, int p=-1) {
    for (int u: to[v]) {
        if (u == p) continue;
        DP d = dp[v] - dp[u].addRoot();
        dp[u] += d.addRoot();
        bfs(u, v);
    }
}

int main(){
    cin.tie(0);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);

    int N;
    cin >> N;

    REP(i, N - 1) {
        int a, b;
        cin >> a >> b;
        a--; b--;
        to[a].pb(b);
        to[b].pb(a);
    }

    dfs(0);
    bfs(0);
    REP(i, N) {
        print(dp[i].addRoot().dp);
    }
}