#include <bits/stdc++.h>
template <typename T>
bool chkmin(T &a, T b) {
  return (b < a) ? a = b, 1 : 0;
}
template <typename T>
bool chkmax(T &a, T b) {
  return (b > a) ? a = b, 1 : 0;
}
long long ksm(long long a, long long b) {
  if (b == 0) return 1;
  long long ns = ksm(a, b >> 1);
  ns = ns * ns % 998244353;
  if (b & 1) ns = ns * a % 998244353;
  return ns;
}
using namespace std;
const int maxn = 100050;
int n;
vector<int> g[maxn];
int cn[maxn];
int ans[maxn];
void dfs(int u, int p, int k) {
  int deg = 0;
  for (int v : g[u]) {
    if (v != p) {
      deg++;
      dfs(v, u, k);
      if (cn[v] % k != 0) {
        cn[v]++;
        deg--;
      }
    }
  }
  cn[u] = deg;
}
int check(int k) {
  for (int i = 1; i <= n; i++) cn[i] = 0;
  dfs(1, 0, k);
  for (int i = 1; i <= n; i++) {
    if (cn[i] % k != 0) {
      return 0;
    }
  }
  return 1;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) g[i].clear();
    for (int i = 1; i < n; i++) {
      int u, v;
      scanf("%d %d", &u, &v);
      g[u].push_back(v);
      g[v].push_back(u);
    }
    ans[1] = ksm(2, n - 1);
    for (int k = 2; k <= n; k++) {
      if ((n - 1) % k != 0) {
        ans[k] = 0;
        continue;
      }
      ans[k] = check(k);
    }
    for (int i = n; i >= 1; i--) {
      for (int j = i + i; j <= n; j += i) {
        ans[i] -= ans[j];
      }
    }
    for (int i = 1; i <= n; i++) {
      printf("%d ", ans[i]);
    }
    printf("\n");
  }
}

### Prompt

Generate 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>
template <typename T>
bool chkmin(T &a, T b) {
  return (b < a) ? a = b, 1 : 0;
}
template <typename T>
bool chkmax(T &a, T b) {
  return (b > a) ? a = b, 1 : 0;
}
long long ksm(long long a, long long b) {
  if (b == 0) return 1;
  long long ns = ksm(a, b >> 1);
  ns = ns * ns % 998244353;
  if (b & 1) ns = ns * a % 998244353;
  return ns;
}
using namespace std;
const int maxn = 100050;
int n;
vector<int> g[maxn];
int cn[maxn];
int ans[maxn];
void dfs(int u, int p, int k) {
  int deg = 0;
  for (int v : g[u]) {
    if (v != p) {
      deg++;
      dfs(v, u, k);
      if (cn[v] % k != 0) {
        cn[v]++;
        deg--;
      }
    }
  }
  cn[u] = deg;
}
int check(int k) {
  for (int i = 1; i <= n; i++) cn[i] = 0;
  dfs(1, 0, k);
  for (int i = 1; i <= n; i++) {
    if (cn[i] % k != 0) {
      return 0;
    }
  }
  return 1;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) g[i].clear();
    for (int i = 1; i < n; i++) {
      int u, v;
      scanf("%d %d", &u, &v);
      g[u].push_back(v);
      g[v].push_back(u);
    }
    ans[1] = ksm(2, n - 1);
    for (int k = 2; k <= n; k++) {
      if ((n - 1) % k != 0) {
        ans[k] = 0;
        continue;
      }
      ans[k] = check(k);
    }
    for (int i = n; i >= 1; i--) {
      for (int j = i + i; j <= n; j += i) {
        ans[i] -= ans[j];
      }
    }
    for (int i = 1; i <= n; i++) {
      printf("%d ", ans[i]);
    }
    printf("\n");
  }
}

```