#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
vector<int> G[100020];
long long num[100010], dp[100010];
int falg;
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
void dfs(int x, int fa, int k) {
  if (falg == 1) return;
  for (int i = 0; i < G[x].size(); i++)
    if (G[x][i] != fa) dfs(G[x][i], x, k);
  if (dp[x] % k == 0)
    dp[fa]++;
  else {
    if (fa) dp[x]++;
    if (dp[x] % k != 0) falg = 1;
  }
  if (falg == 1) return;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, a, b;
    cin >> n;
    for (int i = (0); i < (n + 1); ++i) G[i].clear(), num[i] = dp[i] = 0;
    for (int i = (0); i < (n - 1); ++i) {
      scanf("%d%d", &a, &b);
      G[a].push_back(b);
      G[b].push_back(a);
    }
    num[1] = qpow(2, n - 1);
    for (int i = (2); i < (n); ++i) {
      if ((n - 1) % i == 0) {
        falg = 0;
        dfs(1, 0, i);
        num[i] = 1 - falg;
        for (int i = (0); i < (n + 1); ++i) dp[i] = 0;
      }
    }
    for (int i = n; i >= 1; i--)
      for (int j = 2 * i; j <= n; j += i) num[i] -= num[j];
    for (int i = (1); i < (n + 1); ++i) cout << (num[i] + mod) % mod << ' ';
    cout << '\n';
  }
}

### 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 long long mod = 998244353;
vector<int> G[100020];
long long num[100010], dp[100010];
int falg;
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
void dfs(int x, int fa, int k) {
  if (falg == 1) return;
  for (int i = 0; i < G[x].size(); i++)
    if (G[x][i] != fa) dfs(G[x][i], x, k);
  if (dp[x] % k == 0)
    dp[fa]++;
  else {
    if (fa) dp[x]++;
    if (dp[x] % k != 0) falg = 1;
  }
  if (falg == 1) return;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, a, b;
    cin >> n;
    for (int i = (0); i < (n + 1); ++i) G[i].clear(), num[i] = dp[i] = 0;
    for (int i = (0); i < (n - 1); ++i) {
      scanf("%d%d", &a, &b);
      G[a].push_back(b);
      G[b].push_back(a);
    }
    num[1] = qpow(2, n - 1);
    for (int i = (2); i < (n); ++i) {
      if ((n - 1) % i == 0) {
        falg = 0;
        dfs(1, 0, i);
        num[i] = 1 - falg;
        for (int i = (0); i < (n + 1); ++i) dp[i] = 0;
      }
    }
    for (int i = n; i >= 1; i--)
      for (int j = 2 * i; j <= n; j += i) num[i] -= num[j];
    for (int i = (1); i < (n + 1); ++i) cout << (num[i] + mod) % mod << ' ';
    cout << '\n';
  }
}

```