#include <bits/stdc++.h>
using namespace std;
const int N = 100010, MOD = 998244353;
int Q, n, tot, head[N], f[N];
vector<int> d[N];
struct edge {
  int next, to;
} e[N * 2];
void add(int from, int to) {
  e[++tot] = (edge){head[from], to};
  head[from] = tot;
}
int dfs(int x, int fa, int k) {
  int cnt = 0;
  for (int i = head[x]; ~i; i = e[i].next) {
    int v = e[i].to;
    if (v != fa) {
      int val = dfs(v, x, k);
      if (val == -1) return -1;
      cnt += val;
    }
  }
  if (cnt % k == 0) return 1;
  if (cnt % k == k - 1) return 0;
  return -1;
}
int main() {
  for (int i = 2; i < N; i++)
    for (int j = i; j < N; j += i) d[j].push_back(i);
  scanf("%d", &Q);
  while (Q--) {
    memset(head, -1, sizeof(head));
    memset(f, 0, sizeof(f));
    tot = 0;
    f[1] = 1;
    scanf("%d", &n);
    for (int i = 1, x, y; i < n; i++) {
      scanf("%d%d", &x, &y);
      add(x, y);
      add(y, x);
      f[1] = f[1] * 2 % MOD;
    }
    for (int i = 0; i < d[n - 1].size(); i++)
      if (dfs(1, 0, d[n - 1][i]) == 1) f[d[n - 1][i]] = 1;
    for (int i = n; i >= 1; i--)
      for (int j = i * 2; j <= n; j += i) f[i] -= f[j];
    for (int i = 1; i <= n; i++) cout << f[i] << " ";
    cout << "\n";
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution 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;
const int N = 100010, MOD = 998244353;
int Q, n, tot, head[N], f[N];
vector<int> d[N];
struct edge {
  int next, to;
} e[N * 2];
void add(int from, int to) {
  e[++tot] = (edge){head[from], to};
  head[from] = tot;
}
int dfs(int x, int fa, int k) {
  int cnt = 0;
  for (int i = head[x]; ~i; i = e[i].next) {
    int v = e[i].to;
    if (v != fa) {
      int val = dfs(v, x, k);
      if (val == -1) return -1;
      cnt += val;
    }
  }
  if (cnt % k == 0) return 1;
  if (cnt % k == k - 1) return 0;
  return -1;
}
int main() {
  for (int i = 2; i < N; i++)
    for (int j = i; j < N; j += i) d[j].push_back(i);
  scanf("%d", &Q);
  while (Q--) {
    memset(head, -1, sizeof(head));
    memset(f, 0, sizeof(f));
    tot = 0;
    f[1] = 1;
    scanf("%d", &n);
    for (int i = 1, x, y; i < n; i++) {
      scanf("%d%d", &x, &y);
      add(x, y);
      add(y, x);
      f[1] = f[1] * 2 % MOD;
    }
    for (int i = 0; i < d[n - 1].size(); i++)
      if (dfs(1, 0, d[n - 1][i]) == 1) f[d[n - 1][i]] = 1;
    for (int i = n; i >= 1; i--)
      for (int j = i * 2; j <= n; j += i) f[i] -= f[j];
    for (int i = 1; i <= n; i++) cout << f[i] << " ";
    cout << "\n";
  }
  return 0;
}

```