#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
vector<int> v[100010];
int n, m, now, ans;
int a[100010];
bool flag;
int gcd(int x, int y) { return y == 0 ? x : gcd(y, x % y); }
int dfs(int x, int f) {
  if (flag) return 0;
  int sum = v[x].size();
  for (int i = 0; i < v[x].size(); i++)
    if (v[x][i] != f) sum -= dfs(v[x][i], x);
  if (sum % now && (sum - 1) % now) {
    ans = 1;
    flag = true;
    return 0;
  }
  if (sum % now == 0) {
    ans = gcd(ans, sum);
    return 1;
  }
  ans = gcd(ans, sum - 1);
  return 0;
}
int main() {
  int t, x, y;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (int i = 1; i < n; i++) {
      scanf("%d%d", &x, &y);
      v[x].push_back(y);
      v[y].push_back(x);
    }
    m = n - 1;
    for (now = 2; now <= sqrt(m); now++)
      if (m % now == 0) {
        ans = 0;
        flag = 0;
        dfs(1, -1);
        a[ans] = 1;
        while (m % now == 0) m /= now;
      }
    if (m > 1) {
      ans = 0;
      flag = 0;
      now = m;
      dfs(1, -1);
      a[ans] = 1;
    }
    a[1] = 1;
    for (int i = 1; i < n; i++) a[1] = a[1] * 2 % mod;
    for (int i = 2; i < n; i++) a[1] = (a[1] - a[i] + mod) % mod;
    for (int i = 1; i <= n; i++) printf("%d ", a[i]);
    printf("\n");
    for (int i = 1; i <= n; i++) v[i].clear(), a[i] = 0;
  }
}

### Prompt

Generate 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 mod = 998244353;
vector<int> v[100010];
int n, m, now, ans;
int a[100010];
bool flag;
int gcd(int x, int y) { return y == 0 ? x : gcd(y, x % y); }
int dfs(int x, int f) {
  if (flag) return 0;
  int sum = v[x].size();
  for (int i = 0; i < v[x].size(); i++)
    if (v[x][i] != f) sum -= dfs(v[x][i], x);
  if (sum % now && (sum - 1) % now) {
    ans = 1;
    flag = true;
    return 0;
  }
  if (sum % now == 0) {
    ans = gcd(ans, sum);
    return 1;
  }
  ans = gcd(ans, sum - 1);
  return 0;
}
int main() {
  int t, x, y;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (int i = 1; i < n; i++) {
      scanf("%d%d", &x, &y);
      v[x].push_back(y);
      v[y].push_back(x);
    }
    m = n - 1;
    for (now = 2; now <= sqrt(m); now++)
      if (m % now == 0) {
        ans = 0;
        flag = 0;
        dfs(1, -1);
        a[ans] = 1;
        while (m % now == 0) m /= now;
      }
    if (m > 1) {
      ans = 0;
      flag = 0;
      now = m;
      dfs(1, -1);
      a[ans] = 1;
    }
    a[1] = 1;
    for (int i = 1; i < n; i++) a[1] = a[1] * 2 % mod;
    for (int i = 2; i < n; i++) a[1] = (a[1] - a[i] + mod) % mod;
    for (int i = 1; i <= n; i++) printf("%d ", a[i]);
    printf("\n");
    for (int i = 1; i <= n; i++) v[i].clear(), a[i] = 0;
  }
}

```