#include <bits/stdc++.h>
using namespace std;
int n, a, b, g;
int res[1000007];
struct wierzch {
  list<int> l;
};
wierzch tab[1000007];
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int testuj(int nr, int o, int k) {
  int d = 0;
  for (auto el : tab[nr].l)
    if (el != o) {
      int x = testuj(el, nr, k);
      if (x == -1) return -1;
      d += x;
    }
  if (d % k == 0) {
    g = gcd(d, g);
    return 1;
  }
  if ((d + 1) % k == 0) {
    g = gcd(d + 1, g);
    return 0;
  }
  return -1;
}
int solve() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) tab[i].l.clear();
  for (int i = 0; i < n - 1; i++) {
    scanf("%d %d", &a, &b);
    tab[a].l.push_back(b);
    tab[b].l.push_back(a);
  }
  int r = 1;
  for (int i = 0; i < n - 1; i++) r = r * 2 % 998244353;
  for (int k = 2; k <= n; k++) {
    g = 0;
    if ((n - 1) % k == 0 && testuj(1, 1, k) == 1 && g == k)
      res[k] = 1;
    else
      res[k] = 0;
    r -= res[k];
  }
  res[1] = (r + 998244353) % 998244353;
  for (int i = 1; i <= n; i++) printf("%d ", res[i]);
  printf("\n");
  return 0;
}
int main() {
  int _tests;
  scanf("%d", &_tests);
  for (int _test = 1; _test <= _tests; _test++) solve();
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

You are given a tree with n nodes. As a reminder, a tree is a connected undirected graph without cycles.

Let a_1, a_2, …, a_n be a sequence of integers. Perform the following operation exactly n times: 

  * Select an unerased node u. Assign a_u := number of unerased nodes adjacent to u. Then, erase the node u along with all edges that have it as an endpoint. 



For each integer k from 1 to n, find the number, modulo 998 244 353, of different sequences a_1, a_2, …, a_n that satisfy the following conditions:

  * it is possible to obtain a by performing the aforementioned operations exactly n times in some order. 
  * \operatorname{gcd}(a_1, a_2, …, a_n) = k. Here, \operatorname{gcd} means the greatest common divisor of the elements in a. 

Input

The first line contains a single integer t (1 ≤ t ≤ 10 000) — the number of test cases.

The first line of each test case contains a single integer n (2 ≤ n ≤ 10^5).

Each of the next n - 1 lines contains two integers u and v (1 ≤ u, v ≤ n) indicating there is an edge between vertices u and v. It is guaranteed that the given edges form a tree.

It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.

Output

For each test case, print n integers in a single line, where for each k from 1 to n, the k-th integer denotes the answer when \operatorname{gcd} equals to k.

Example

Input


2
3
2 1
1 3
2
1 2


Output


3 1 0
2 0

Note

In the first test case, 

<image>

  * If we delete the nodes in order 1 → 2 → 3 or 1 → 3 → 2, then the obtained sequence will be a = [2, 0, 0] which has \operatorname{gcd} equals to 2. 
  * If we delete the nodes in order 2 → 1 → 3, then the obtained sequence will be a = [1, 1, 0] which has \operatorname{gcd} equals to 1. 
  * If we delete the nodes in order 3 → 1 → 2, then the obtained sequence will be a = [1, 0, 1] which has \operatorname{gcd} equals to 1. 
  * If we delete the nodes in order 2 → 3 → 1 or 3 → 2 → 1, then the obtained sequence will be a = [0, 1, 1] which has \operatorname{gcd} equals to 1. 



Note that here we are counting the number of different sequences, not the number of different orders of deleting nodes.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, a, b, g;
int res[1000007];
struct wierzch {
  list<int> l;
};
wierzch tab[1000007];
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int testuj(int nr, int o, int k) {
  int d = 0;
  for (auto el : tab[nr].l)
    if (el != o) {
      int x = testuj(el, nr, k);
      if (x == -1) return -1;
      d += x;
    }
  if (d % k == 0) {
    g = gcd(d, g);
    return 1;
  }
  if ((d + 1) % k == 0) {
    g = gcd(d + 1, g);
    return 0;
  }
  return -1;
}
int solve() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) tab[i].l.clear();
  for (int i = 0; i < n - 1; i++) {
    scanf("%d %d", &a, &b);
    tab[a].l.push_back(b);
    tab[b].l.push_back(a);
  }
  int r = 1;
  for (int i = 0; i < n - 1; i++) r = r * 2 % 998244353;
  for (int k = 2; k <= n; k++) {
    g = 0;
    if ((n - 1) % k == 0 && testuj(1, 1, k) == 1 && g == k)
      res[k] = 1;
    else
      res[k] = 0;
    r -= res[k];
  }
  res[1] = (r + 998244353) % 998244353;
  for (int i = 1; i <= n; i++) printf("%d ", res[i]);
  printf("\n");
  return 0;
}
int main() {
  int _tests;
  scanf("%d", &_tests);
  for (int _test = 1; _test <= _tests; _test++) solve();
  return 0;
}

```