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.

Examples

Input

3
1 2
1 3


Output

2
1
1


Input

2
1 2


Output

1
1


Input

5
1 2
2 3
3 4
3 5


Output

2
8
12
3
3


Input

8
1 2
2 3
3 4
3 5
3 6
6 7
6 8


Output

40
280
840
120
120
504
72
72
#include <bits/stdc++.h>
using namespace std;
static auto faster_iostream = []() { ios::sync_with_stdio(0); cin.tie(0); return 0; }();void print(){cout<<'\n';}template<class T,class ...U>void print(T x, U ...y){cout<<x<<' ';print(y...);}
typedef long long ll; typedef long double db;
const int N = 2e5 + 5, M = 1e9 + 7;
ll qpow(ll x, ll n) { ll res = 1; for (; n; n >>= 1, x = x * x % M) if (n & 1) res = res * x % M; return res; }

ll fac[N];
ll comb(ll n, ll m) {
  return fac[n] * qpow(fac[m] * fac[n-m] % M, M - 2) % M;
}

ll sum[N], siz[N], ans[N];
vector<int> G[N];

void dfs(int u, int f) {
  sum[u] = 1;
  for (int v : G[u]) {
    if (v == f) continue;
    dfs(v, u);
    siz[u] += siz[v];
    sum[u] = sum[u] * comb(siz[u], siz[v]) % M * sum[v] % M;
  }
  siz[u]++;
}

int n;

void sol(int u, int f) {
  ans[u] = sum[u];
  for (int v : G[u]) {
    if (v == f) continue;
    ll tsum = sum[v];
    sum[v] = sum[v] * comb(n - 1, siz[v] - 1) % M * sum[u] % M * qpow(comb(n - 1, siz[v]) * sum[v] % M, M - 2) % M;
    sol(v, u);
    sum[v] = tsum;
  }
}

int main() {
  fac[0] = fac[1] = 1;
  for (int i = 2; i < N; i++) {
    fac[i] = fac[i-1] * i % M;
  }
  int u, v;
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 0);
  sol(1, 0);
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << '\n';
  }
}
