#include <bits/stdc++.h>
using namespace std;
int T, n, k, dp[100010], fl, ans[100010];
vector<int> g[100010];
const int mod = 998244353;
void dfs(int x, int f) {
  int s = 0;
  for (int i = 0, v; i < g[x].size(); i++) {
    v = g[x][i];
    if (v == f) continue;
    dfs(v, x);
    s += dp[v];
  }
  dp[x] = 1;
  if (f && s % k) s++, dp[x] = 0;
  if (s % k) fl = 1;
}
const int N = 1e5;
int mu[N + 10], f[N + 10], pri[N + 10], cn;
void sieve() {
  mu[1] = 1;
  for (int i = 2; i <= N; i++) {
    if (!f[i]) pri[++cn] = i, mu[i] = -1;
    for (int j = 1; j <= cn && i * pri[j] <= N; j++) {
      f[i * pri[j]] = 1;
      if (!(i % pri[j])) break;
      mu[i * pri[j]] = -mu[i];
    }
  }
}
void sol() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) g[i].clear(), ans[i] = dp[i] = f[i] = 0;
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  f[1] = 1;
  for (int i = 1; i < n; i++) (f[1] *= 2) %= mod;
  for (k = 2; k < n; k++)
    if (!((n - 1) % k)) {
      fl = 0;
      dfs(1, 0);
      if (!fl) f[k] = 1;
    }
  for (int i = 1; i <= n; i++)
    for (int j = 1; i * j <= n; j++) (ans[i] += f[i * j] * mu[j]) %= mod;
  for (int i = 1; i <= n; i++) printf("%d ", (ans[i] + mod) % mod);
  printf("\n");
}
int main() {
  sieve();
  int T;
  scanf("%d", &T);
  while (T--) sol();
  return 0;
}

### Prompt

Please formulate 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;
int T, n, k, dp[100010], fl, ans[100010];
vector<int> g[100010];
const int mod = 998244353;
void dfs(int x, int f) {
  int s = 0;
  for (int i = 0, v; i < g[x].size(); i++) {
    v = g[x][i];
    if (v == f) continue;
    dfs(v, x);
    s += dp[v];
  }
  dp[x] = 1;
  if (f && s % k) s++, dp[x] = 0;
  if (s % k) fl = 1;
}
const int N = 1e5;
int mu[N + 10], f[N + 10], pri[N + 10], cn;
void sieve() {
  mu[1] = 1;
  for (int i = 2; i <= N; i++) {
    if (!f[i]) pri[++cn] = i, mu[i] = -1;
    for (int j = 1; j <= cn && i * pri[j] <= N; j++) {
      f[i * pri[j]] = 1;
      if (!(i % pri[j])) break;
      mu[i * pri[j]] = -mu[i];
    }
  }
}
void sol() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) g[i].clear(), ans[i] = dp[i] = f[i] = 0;
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  f[1] = 1;
  for (int i = 1; i < n; i++) (f[1] *= 2) %= mod;
  for (k = 2; k < n; k++)
    if (!((n - 1) % k)) {
      fl = 0;
      dfs(1, 0);
      if (!fl) f[k] = 1;
    }
  for (int i = 1; i <= n; i++)
    for (int j = 1; i * j <= n; j++) (ans[i] += f[i * j] * mu[j]) %= mod;
  for (int i = 1; i <= n; i++) printf("%d ", (ans[i] + mod) % mod);
  printf("\n");
}
int main() {
  sieve();
  int T;
  scanf("%d", &T);
  while (T--) sol();
  return 0;
}

```