#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
const unsigned gen_seed =
    std::chrono::system_clock::now().time_since_epoch().count();
std::mt19937_64 gen(gen_seed);
std::vector<std::pair<int, int>> ord;
vector<set<int>> g;
std::vector<int> deg;
void go(int v) {
  if (deg[v] == 1) {
    int to = *g[v].begin();
    ord.push_back(make_pair(v, to));
    g[to].erase(v);
    g[v].erase(to);
    deg[to]--;
    deg[v]--;
    go(to);
  }
}
int main() {
  int t;
  scanf("%d", &t);
  for (int afaf = 0; afaf < t; afaf++) {
    int n;
    scanf("%d", &n);
    g = vector<set<int>>(n);
    deg = std::vector<int>(n, 0);
    for (int i = 0; i < n - 1; i++) {
      int u, v;
      scanf("%d %d", &u, &v);
      u--;
      v--;
      g[u].insert(v);
      g[v].insert(u);
      deg[u]++;
      deg[v]++;
    }
    ord = std::vector<std::pair<int, int>>();
    for (int i = 0; i < n; i++) go(i);
    std::vector<int> ans(n + 1, 0);
    long long lft = 1;
    for (int i = 0; i < n - 1; i++) lft = lft * 2 % mod;
    for (int d = n - 1; d >= 2; d--) {
      if ((n - 1) % d == 0) {
        std::vector<int> val(n, 0);
        for (auto x : ord) {
          if (val[x.first] % d == 0) {
            val[x.second]++;
          } else
            val[x.first]++;
        }
        int can = 1;
        for (int i = 0; i < n; i++)
          if (val[i] % d != 0) can = 0;
        for (int d2 = 2 * d; d2 < n; d2 += d) {
          if (ans[d2] == 1) can = 0;
        }
        if (can == 1) {
          ans[d] = 1;
          lft = (lft + mod - 1) % mod;
        }
      }
    }
    ans[1] = lft;
    for (int i = 1; i < n + 1; i++) {
      printf("%d", ans[i]);
      if (i != n)
        printf(" ");
      else
        printf("\n");
    }
  }
}

### 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>
using namespace std;
const long long mod = 998244353;
const unsigned gen_seed =
    std::chrono::system_clock::now().time_since_epoch().count();
std::mt19937_64 gen(gen_seed);
std::vector<std::pair<int, int>> ord;
vector<set<int>> g;
std::vector<int> deg;
void go(int v) {
  if (deg[v] == 1) {
    int to = *g[v].begin();
    ord.push_back(make_pair(v, to));
    g[to].erase(v);
    g[v].erase(to);
    deg[to]--;
    deg[v]--;
    go(to);
  }
}
int main() {
  int t;
  scanf("%d", &t);
  for (int afaf = 0; afaf < t; afaf++) {
    int n;
    scanf("%d", &n);
    g = vector<set<int>>(n);
    deg = std::vector<int>(n, 0);
    for (int i = 0; i < n - 1; i++) {
      int u, v;
      scanf("%d %d", &u, &v);
      u--;
      v--;
      g[u].insert(v);
      g[v].insert(u);
      deg[u]++;
      deg[v]++;
    }
    ord = std::vector<std::pair<int, int>>();
    for (int i = 0; i < n; i++) go(i);
    std::vector<int> ans(n + 1, 0);
    long long lft = 1;
    for (int i = 0; i < n - 1; i++) lft = lft * 2 % mod;
    for (int d = n - 1; d >= 2; d--) {
      if ((n - 1) % d == 0) {
        std::vector<int> val(n, 0);
        for (auto x : ord) {
          if (val[x.first] % d == 0) {
            val[x.second]++;
          } else
            val[x.first]++;
        }
        int can = 1;
        for (int i = 0; i < n; i++)
          if (val[i] % d != 0) can = 0;
        for (int d2 = 2 * d; d2 < n; d2 += d) {
          if (ans[d2] == 1) can = 0;
        }
        if (can == 1) {
          ans[d] = 1;
          lft = (lft + mod - 1) % mod;
        }
      }
    }
    ans[1] = lft;
    for (int i = 1; i < n + 1; i++) {
      printf("%d", ans[i]);
      if (i != n)
        printf(" ");
      else
        printf("\n");
    }
  }
}

```