#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &a) {
  T w = 1;
  a = 0;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') w = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar())
    a = (a << 3) + (a << 1) + (ch ^ 48);
  a *= w;
}
template <typename T>
inline void ckmax(T &a, T b) {
  a = a > b ? a : b;
}
template <typename T>
inline void ckmin(T &a, T b) {
  a = a < b ? a : b;
}
const int N = 1e5 + 10;
const int Mod = 998244353;
vector<int> to[N];
int f[N], g[N];
inline int Sub(int x, int y) { return (x - y) < 0 ? (x + Mod - y) : (x - y); }
inline int Mul(int x, int y) {
  return (1ll * x * y) >= Mod ? (1ll * x * y % Mod) : (x * y);
}
int qpow(int a, int b) {
  int base = a, ans = 1;
  while (b) {
    if (b & 1) ans = Mul(ans, base);
    base = Mul(base, base);
    b >>= 1;
  }
  return ans;
}
int fl;
int dfs(int u, int fa, int k) {
  int cnt = 0;
  for (auto v : to[u]) {
    if (v == fa) continue;
    cnt += dfs(v, u, k);
  }
  if (!(cnt % k)) return 1;
  if (!((cnt + 1) % k)) return 0;
  fl = 0;
  return 0;
}
int main() {
  int cases;
  read(cases);
  while (cases--) {
    int n;
    read(n);
    memset(f, 0, sizeof f);
    for (int i = 1, u, v; i < n; ++i) {
      read(u);
      read(v);
      to[u].push_back(v);
      to[v].push_back(u);
    }
    f[1] = qpow(2, n - 1);
    for (int i = 2; i < n; ++i) {
      if (!((n - 1) % i)) {
        fl = 1;
        f[i] = dfs(1, 0, i) && fl;
      }
    }
    for (int i = n - 1; i >= 1; --i) {
      for (int j = i + i; j < n; j += i) f[i] = Sub(f[i], f[j]);
    }
    for (int i = 1; i <= n; ++i) printf("%d ", f[i]);
    puts("");
    for (int i = 1; i <= n; ++i) to[i].clear();
  }
}

### Prompt

Your task is to create 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;
template <typename T>
inline void read(T &a) {
  T w = 1;
  a = 0;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') w = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar())
    a = (a << 3) + (a << 1) + (ch ^ 48);
  a *= w;
}
template <typename T>
inline void ckmax(T &a, T b) {
  a = a > b ? a : b;
}
template <typename T>
inline void ckmin(T &a, T b) {
  a = a < b ? a : b;
}
const int N = 1e5 + 10;
const int Mod = 998244353;
vector<int> to[N];
int f[N], g[N];
inline int Sub(int x, int y) { return (x - y) < 0 ? (x + Mod - y) : (x - y); }
inline int Mul(int x, int y) {
  return (1ll * x * y) >= Mod ? (1ll * x * y % Mod) : (x * y);
}
int qpow(int a, int b) {
  int base = a, ans = 1;
  while (b) {
    if (b & 1) ans = Mul(ans, base);
    base = Mul(base, base);
    b >>= 1;
  }
  return ans;
}
int fl;
int dfs(int u, int fa, int k) {
  int cnt = 0;
  for (auto v : to[u]) {
    if (v == fa) continue;
    cnt += dfs(v, u, k);
  }
  if (!(cnt % k)) return 1;
  if (!((cnt + 1) % k)) return 0;
  fl = 0;
  return 0;
}
int main() {
  int cases;
  read(cases);
  while (cases--) {
    int n;
    read(n);
    memset(f, 0, sizeof f);
    for (int i = 1, u, v; i < n; ++i) {
      read(u);
      read(v);
      to[u].push_back(v);
      to[v].push_back(u);
    }
    f[1] = qpow(2, n - 1);
    for (int i = 2; i < n; ++i) {
      if (!((n - 1) % i)) {
        fl = 1;
        f[i] = dfs(1, 0, i) && fl;
      }
    }
    for (int i = n - 1; i >= 1; --i) {
      for (int j = i + i; j < n; j += i) f[i] = Sub(f[i], f[j]);
    }
    for (int i = 1; i <= n; ++i) printf("%d ", f[i]);
    puts("");
    for (int i = 1; i <= n; ++i) to[i].clear();
  }
}

```