#include <bits/stdc++.h>
using namespace std;
int t, n;
vector<int> g[100100];
long long ans[100100];
bool extra[100100];
bool poss(int c, int l, int k) {
  int cv = 0;
  for (int i : g[c]) {
    if (i == l) continue;
    if (!poss(i, c, k)) return false;
    cv += extra[i];
  }
  if (cv % k == 0) {
    extra[c] = true;
    return true;
  }
  if ((cv + 1) % k == 0) {
    extra[c] = false;
    return true;
  }
  return false;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    long long ctot = 1;
    for (int i = 1; i < n; i++) {
      int n1, n2;
      scanf(" %d%d", &n1, &n2);
      g[n1].push_back(n2);
      g[n2].push_back(n1);
      ctot *= 2;
      ctot %= 998244353;
    }
    ans[1] = ctot;
    for (int i = n; i > 1; i--) {
      ans[i] = 0;
      if ((n - 1) % i) continue;
      for (int j = i * 2; j <= n; j += i)
        if (ans[j]) ans[i] = 1;
      if (ans[i]) continue;
      for (int j = 1; j <= n; j++) extra[j] = false;
      bool cv = poss(1, -1, i);
      if (cv && extra[1]) ans[i] = 1;
    }
    for (int i = n; i >= 1; i--) {
      for (int j = i * 2; j <= n; j += i) ans[i] -= ans[j];
      ans[i] += 998244353;
      ans[i] %= 998244353;
    }
    for (int i = 1; i <= n; i++) printf("%lld ", ans[i]);
    printf("\n");
    for (int i = 1; i <= n; i++) g[i].clear();
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 t, n;
vector<int> g[100100];
long long ans[100100];
bool extra[100100];
bool poss(int c, int l, int k) {
  int cv = 0;
  for (int i : g[c]) {
    if (i == l) continue;
    if (!poss(i, c, k)) return false;
    cv += extra[i];
  }
  if (cv % k == 0) {
    extra[c] = true;
    return true;
  }
  if ((cv + 1) % k == 0) {
    extra[c] = false;
    return true;
  }
  return false;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    long long ctot = 1;
    for (int i = 1; i < n; i++) {
      int n1, n2;
      scanf(" %d%d", &n1, &n2);
      g[n1].push_back(n2);
      g[n2].push_back(n1);
      ctot *= 2;
      ctot %= 998244353;
    }
    ans[1] = ctot;
    for (int i = n; i > 1; i--) {
      ans[i] = 0;
      if ((n - 1) % i) continue;
      for (int j = i * 2; j <= n; j += i)
        if (ans[j]) ans[i] = 1;
      if (ans[i]) continue;
      for (int j = 1; j <= n; j++) extra[j] = false;
      bool cv = poss(1, -1, i);
      if (cv && extra[1]) ans[i] = 1;
    }
    for (int i = n; i >= 1; i--) {
      for (int j = i * 2; j <= n; j += i) ans[i] -= ans[j];
      ans[i] += 998244353;
      ans[i] %= 998244353;
    }
    for (int i = 1; i <= n; i++) printf("%lld ", ans[i]);
    printf("\n");
    for (int i = 1; i <= n; i++) g[i].clear();
  }
}

```