#include <bits/stdc++.h>
using namespace std;
int T, n, ans[200005], val[200005];
int cnt, pri[200005], vis[200005], mu[200005];
vector<int> g[200005];
void init(int lim) {
  mu[1] = 1;
  for (int i = 2; i <= lim; ++i) {
    if (!vis[i]) pri[++cnt] = i, mu[i] = -1;
    for (int j = 1; j <= cnt && pri[j] * i <= lim; ++j) {
      vis[pri[j] * i] = 1;
      if (i % pri[j] == 0) {
        mu[i * pri[j]] = 0;
        break;
      }
      mu[pri[j] * i] = -mu[i];
    }
  }
}
int ksm(int x, int y) {
  int num = 1;
  while (y) {
    if (y & 1) num = 1LL * num * x % 998244353;
    x = 1LL * x * x % 998244353, y >>= 1;
  }
  return num;
}
bool dfs(int x, int fa, int k) {
  val[x] = 0;
  int flag = 1, num = 0;
  for (int to : g[x])
    if (to != fa) {
      if (!dfs(to, x, k)) flag = 0;
      if (!val[to]) num++;
    }
  if (num % k == 0)
    val[x] = 0;
  else if ((num + 1) % k == 0)
    val[x] = 1;
  else
    flag = 0;
  return flag;
}
int main() {
  scanf("%d", &T);
  init(1e5);
  while (T--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
      ans[i] = 0;
      g[i].clear();
    }
    for (int i = 1; i < n; ++i) {
      int x, y;
      scanf("%d%d", &x, &y);
      g[x].push_back(y), g[y].push_back(x);
    }
    for (int i = 1; i * i <= (n - 1); ++i)
      if ((n - 1) % i == 0) {
        ans[i] = dfs(1, 0, i);
        if (i * i != n - 1) ans[(n - 1) / i] = dfs(1, 0, (n - 1) / i);
      }
    ans[1] = ksm(2, n - 1);
    for (int i = 1; i <= n; ++i) {
      for (int j = 2; j * i <= n; ++j)
        ans[i] = (ans[i] + mu[j] * ans[i * j] + 998244353) % 998244353;
      printf("%d ", ans[i]);
    }
    printf("\n");
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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, ans[200005], val[200005];
int cnt, pri[200005], vis[200005], mu[200005];
vector<int> g[200005];
void init(int lim) {
  mu[1] = 1;
  for (int i = 2; i <= lim; ++i) {
    if (!vis[i]) pri[++cnt] = i, mu[i] = -1;
    for (int j = 1; j <= cnt && pri[j] * i <= lim; ++j) {
      vis[pri[j] * i] = 1;
      if (i % pri[j] == 0) {
        mu[i * pri[j]] = 0;
        break;
      }
      mu[pri[j] * i] = -mu[i];
    }
  }
}
int ksm(int x, int y) {
  int num = 1;
  while (y) {
    if (y & 1) num = 1LL * num * x % 998244353;
    x = 1LL * x * x % 998244353, y >>= 1;
  }
  return num;
}
bool dfs(int x, int fa, int k) {
  val[x] = 0;
  int flag = 1, num = 0;
  for (int to : g[x])
    if (to != fa) {
      if (!dfs(to, x, k)) flag = 0;
      if (!val[to]) num++;
    }
  if (num % k == 0)
    val[x] = 0;
  else if ((num + 1) % k == 0)
    val[x] = 1;
  else
    flag = 0;
  return flag;
}
int main() {
  scanf("%d", &T);
  init(1e5);
  while (T--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
      ans[i] = 0;
      g[i].clear();
    }
    for (int i = 1; i < n; ++i) {
      int x, y;
      scanf("%d%d", &x, &y);
      g[x].push_back(y), g[y].push_back(x);
    }
    for (int i = 1; i * i <= (n - 1); ++i)
      if ((n - 1) % i == 0) {
        ans[i] = dfs(1, 0, i);
        if (i * i != n - 1) ans[(n - 1) / i] = dfs(1, 0, (n - 1) / i);
      }
    ans[1] = ksm(2, n - 1);
    for (int i = 1; i <= n; ++i) {
      for (int j = 2; j * i <= n; ++j)
        ans[i] = (ans[i] + mu[j] * ans[i * j] + 998244353) % 998244353;
      printf("%d ", ans[i]);
    }
    printf("\n");
  }
  return 0;
}

```