#include <bits/stdc++.h>
using namespace std;
inline bool isdigit(char& ch) { return ch >= '0' && ch <= '9'; }
template <class T>
void read(T& x) {
  x = 0;
  long long f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
  x *= f;
}
template <class T>
inline void write(T x) {
  if (x == 0) {
    putchar('0');
    return;
  }
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  int _stk[65], _top = 0;
  for (; x; x /= 10) _stk[++_top] = x % 10 + 48;
  for (; _top; _top--) putchar(_stk[_top]);
}
const int MAXN = 1e5 + 10;
const int mod = 998244353;
int f[MAXN], ans[MAXN], n, flag;
vector<int> g[MAXN];
void dfs(int u, int fa, int d) {
  if (!flag) return;
  for (auto v : g[u]) {
    if (v == fa) continue;
    dfs(v, u, d);
  }
  if (f[u] % d == 0)
    f[fa]++;
  else {
    if (fa) f[u]++;
    if (f[u] % d != 0) flag = 0;
  }
  if (!flag) return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int T;
  cin >> T;
  while (T--) {
    cin >> n;
    for (int i = (int)(1); i < (int)(n); i++) {
      int u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    ans[1] = 1;
    for (int i = (int)(1); i < (int)(n); i++)
      ans[1] = (1ll * ans[1] + ans[1]) % mod;
    for (int d = (int)(2); d < (int)(n + 1); d++) {
      if ((n - 1) % d == 0) {
        flag = 1;
        dfs(1, 0, d);
        ans[d] = flag;
        for (int i = (int)(1); i < (int)(n + 1); i++) f[i] = 0;
      }
    }
    for (int i = n; i >= 1; --i) {
      for (int j = i + i; j <= n; j += i) ans[i] = (ans[i] - ans[j]) % mod;
    }
    for (int i = (int)(1); i < (int)(n + 1); i++) {
      ans[i] = (ans[i] % mod + mod) % mod;
      cout << ans[i] << ' ';
    }
    cout << '\n';
    for (int i = (int)(1); i < (int)(n + 1); i++) g[i].clear(), ans[i] = 0;
  }
  return 0;
}

### 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;
inline bool isdigit(char& ch) { return ch >= '0' && ch <= '9'; }
template <class T>
void read(T& x) {
  x = 0;
  long long f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
  x *= f;
}
template <class T>
inline void write(T x) {
  if (x == 0) {
    putchar('0');
    return;
  }
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  int _stk[65], _top = 0;
  for (; x; x /= 10) _stk[++_top] = x % 10 + 48;
  for (; _top; _top--) putchar(_stk[_top]);
}
const int MAXN = 1e5 + 10;
const int mod = 998244353;
int f[MAXN], ans[MAXN], n, flag;
vector<int> g[MAXN];
void dfs(int u, int fa, int d) {
  if (!flag) return;
  for (auto v : g[u]) {
    if (v == fa) continue;
    dfs(v, u, d);
  }
  if (f[u] % d == 0)
    f[fa]++;
  else {
    if (fa) f[u]++;
    if (f[u] % d != 0) flag = 0;
  }
  if (!flag) return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int T;
  cin >> T;
  while (T--) {
    cin >> n;
    for (int i = (int)(1); i < (int)(n); i++) {
      int u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    ans[1] = 1;
    for (int i = (int)(1); i < (int)(n); i++)
      ans[1] = (1ll * ans[1] + ans[1]) % mod;
    for (int d = (int)(2); d < (int)(n + 1); d++) {
      if ((n - 1) % d == 0) {
        flag = 1;
        dfs(1, 0, d);
        ans[d] = flag;
        for (int i = (int)(1); i < (int)(n + 1); i++) f[i] = 0;
      }
    }
    for (int i = n; i >= 1; --i) {
      for (int j = i + i; j <= n; j += i) ans[i] = (ans[i] - ans[j]) % mod;
    }
    for (int i = (int)(1); i < (int)(n + 1); i++) {
      ans[i] = (ans[i] % mod + mod) % mod;
      cout << ans[i] << ' ';
    }
    cout << '\n';
    for (int i = (int)(1); i < (int)(n + 1); i++) g[i].clear(), ans[i] = 0;
  }
  return 0;
}

```