#include <bits/stdc++.h>
const int mod = 998244353;
using std::vector;
int ans[100000 + 5];
vector<int> e[100000 + 5];
int qpow(int bsc, int y) {
  int ret = 1;
  while (y) {
    if (y & 1) ret = (long long)ret * bsc % mod;
    bsc = (long long)bsc * bsc % mod;
    y >>= 1;
  }
  return ret;
}
bool flag;
int d;
int dfs(int u, int f) {
  int cnt = 0;
  for (int v : e[u]) {
    if (v == f) continue;
    int sum = dfs(v, u);
    if (!flag) return 0;
    if (sum % d && (sum + 1) % d) {
      flag = false;
      return 0;
    }
    cnt += sum + 1;
  }
  return cnt;
}
bool check(int _d) {
  flag = true;
  d = _d;
  dfs(1, 0);
  return flag;
}
void solve() {
  int n;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    e[u].push_back(v);
    e[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) ans[i] = 0;
  int cnto = 0;
  vector<int> fs;
  for (int i = 2; i * i <= n - 1; i++)
    if ((n - 1) % i == 0) {
      fs.push_back(i);
      if (i * i != n - 1) fs.push_back((n - 1) / i);
    }
  if (n - 1 != 1) fs.push_back(n - 1);
  for (int d : fs) {
    if (check(d)) {
      cnto++;
      ans[d] = 1;
    }
  }
  for (int x : fs) {
    if (!ans[x]) continue;
    for (int y : fs) {
      if (!ans[y]) continue;
      if (x != y && x % y == 0) {
        ans[y] = 0;
        cnto--;
      }
    }
  }
  ans[1] = (qpow(2, n - 1) + mod - cnto) % mod;
  for (int i = 1; i <= n; i++) printf("%d%c", ans[i], " \n"[i == n]);
  for (int i = 1; i <= n; i++) {
    e[i].clear();
  }
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) solve();
}

### 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>
const int mod = 998244353;
using std::vector;
int ans[100000 + 5];
vector<int> e[100000 + 5];
int qpow(int bsc, int y) {
  int ret = 1;
  while (y) {
    if (y & 1) ret = (long long)ret * bsc % mod;
    bsc = (long long)bsc * bsc % mod;
    y >>= 1;
  }
  return ret;
}
bool flag;
int d;
int dfs(int u, int f) {
  int cnt = 0;
  for (int v : e[u]) {
    if (v == f) continue;
    int sum = dfs(v, u);
    if (!flag) return 0;
    if (sum % d && (sum + 1) % d) {
      flag = false;
      return 0;
    }
    cnt += sum + 1;
  }
  return cnt;
}
bool check(int _d) {
  flag = true;
  d = _d;
  dfs(1, 0);
  return flag;
}
void solve() {
  int n;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    e[u].push_back(v);
    e[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) ans[i] = 0;
  int cnto = 0;
  vector<int> fs;
  for (int i = 2; i * i <= n - 1; i++)
    if ((n - 1) % i == 0) {
      fs.push_back(i);
      if (i * i != n - 1) fs.push_back((n - 1) / i);
    }
  if (n - 1 != 1) fs.push_back(n - 1);
  for (int d : fs) {
    if (check(d)) {
      cnto++;
      ans[d] = 1;
    }
  }
  for (int x : fs) {
    if (!ans[x]) continue;
    for (int y : fs) {
      if (!ans[y]) continue;
      if (x != y && x % y == 0) {
        ans[y] = 0;
        cnto--;
      }
    }
  }
  ans[1] = (qpow(2, n - 1) + mod - cnto) % mod;
  for (int i = 1; i <= n; i++) printf("%d%c", ans[i], " \n"[i == n]);
  for (int i = 1; i <= n; i++) {
    e[i].clear();
  }
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) solve();
}

```