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;
using ll = long long;
using pii = pair<int, int>;
#define FI(i, a, b) for (int i = (a); i <= (b); ++i)
#define FD(i, b, a) for (int i = (b); i >= (a); --i)
#define DEBUG(x) cerr << #x << ": " << (x) << endl;

const int N = 2e5 + 5, M = 1e9 + 7;
int n, f[N], cnt[N], fac[N];
vector<int> g[N];

constexpr long long inv(long long x, long long m) {
  return x > m ? inv(x % m, m) : x > 1 ? inv(m % x, m) * (m - m / x) % m : x;
}

void dfs1(int u, int p) {
  cnt[u] = 1, f[u] = 1;
  for (auto v: g[u]) {
    if (v == p) continue;
    dfs1(v, u);
    cnt[u] += cnt[v];
    f[u] = 1ll * f[u] * f[v] % M * inv(fac[cnt[v]], M) % M;
  }
  f[u] = 1ll * f[u] * fac[cnt[u] - 1] % M;
}

void dfs2(int u, int p) {
  if (u != p) {
    f[u] = 1ll * f[p] * cnt[u] % M * inv(n - cnt[u], M) % M;
  }  
  for (auto v: g[u]) {
    if (v == p) continue;
    dfs2(v, u);
  }
}

int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = 1ll * fac[i - 1] * i % M; 
  dfs1(1, 1);
  dfs2(1, 1);
  for (int i = 1; i <= n; ++i) printf("%d\n", f[i]);
}