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<cstdio>
#include<vector>

using namespace std;

int n;
vector<int> v[200005];
int cnt[200005];
int s[200005];
int f[200005];
int ans[200005];
const int MOD = 1000000007;

int pow(int x, int y) {
  if (y == 0) {
    return 1;
  }
  if (y % 2 == 0) {
    return pow(1ll * x * x % MOD, y / 2);
  } else {
    return 1ll * x * pow(x, y - 1) % MOD;
  }
}

int inv(int x) {
  return pow(x, MOD - 2);
}

int nCr(int n, int r) {
  return 1ll * f[n] * inv(f[r]) % MOD * inv(f[n - r]) % MOD;
}

void dfs(int x, int y) {
  for (auto u: v[x]) {
    if (u != y) {
      dfs(u, x);
    }
  }
  cnt[x] = 1;
  s[x] = 1;
  for (auto u: v[x]) {
    s[x] += s[u];
  }
  int z = s[x] - 1;
  for (auto u: v[x]) {
    if (u != y) {
      cnt[x] = 1ll * cnt[x] * nCr(z, s[u]) % MOD * cnt[u] % MOD;
      z -= s[u];
    }
  }
}

void dfs2(int x, int y) {
  int z = s[1] - 1;
  ans[x] = 1ll * nCr(z, s[1] - s[x]) * (1ll * ans[y] * inv(1ll * nCr(z, s[x]) * cnt[x] % MOD) % MOD) % MOD;
  z -= s[1] - s[x];
  for (auto u: v[x]) {
    if (u != y) {
      ans[x] = 1ll * ans[x] * nCr(z, s[u]) % MOD * cnt[u] % MOD;
      z -= s[u];
    }
  }
  for (auto u: v[x]) {
    if (u != y) {
      dfs2(u, x);
    }
  }
}

int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    v[a].push_back(b);
    v[b].push_back(a);
  }
  f[0] = 1;
  for (int i = 1; i <= n; i++) {
    f[i] = 1ll * i * f[i - 1] % MOD;
  }
  dfs(1, 0);
  ans[1] = cnt[1];
  for (auto u: v[1]) {
    dfs2(u, 1);
  }
  for (int i = 1; i <= n; i++) {
    printf("%d\n", ans[i]);
  }
}
