#include <bits/stdc++.h>
using namespace std;
int read() {
  char c = getchar();
  int s = 0;
  int x = 1;
  while (c < '0' || c > '9') {
    if (c == '-') x = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    s = s * 10 + c - '0';
    c = getchar();
  }
  return s * x;
}
const int mod = 998244353;
const int N = 1e5 + 5;
int T, n, x, y, f[N], ans[N];
vector<int> v[N];
int ksm(int x, int k) {
  int base = 1;
  while (k) {
    if (k & 1) base = 1ll * base * x % mod;
    k >>= 1;
    x = 1ll * x * x % mod;
  }
  return base;
}
int gcd(int x, int y) {
  if (!x) return y;
  return gcd(y % x, x);
}
void dfs(int u, int fa, int p) {
  int siz = 0;
  for (int i = 0; i < v[u].size(); i++) {
    if (v[u][i] == fa) continue;
    dfs(v[u][i], u, p);
    siz++;
  }
  if (!siz) {
    f[fa]++;
  } else {
    if (fa) {
      if (f[u] % p == 0)
        f[fa]++;
      else
        f[u]++;
    }
  }
}
int main() {
  T = read();
  while (T--) {
    n = read();
    for (int i = 1; i < n; i++) {
      x = read();
      y = read();
      v[x].push_back(y);
      v[y].push_back(x);
    }
    int tmp = ksm(2, n - 1);
    int pre = 0;
    for (int i = 2; i < n; i++) {
      if ((n - 1) % i == 0) {
        memset(f, 0, (n + 1) * 4);
        dfs(1, 0, i);
        int mi = 0;
        bool flag = 0;
        for (int k = 1; k <= n; k++) {
          mi = gcd(f[k], mi);
          if (f[k] % i) flag = 1;
        }
        if (!flag && mi == i) ans[i] = 1, pre++;
      }
    }
    cout << (tmp - pre + mod) % mod << " ";
    for (int i = 2; i <= n; i++) {
      cout << ans[i] << " ";
    }
    memset(ans, 0, (n + 1) * 4);
    for (int i = 1; i <= n; i++) v[i].clear();
    puts("");
  }
  return 0;
}

### 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>
using namespace std;
int read() {
  char c = getchar();
  int s = 0;
  int x = 1;
  while (c < '0' || c > '9') {
    if (c == '-') x = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    s = s * 10 + c - '0';
    c = getchar();
  }
  return s * x;
}
const int mod = 998244353;
const int N = 1e5 + 5;
int T, n, x, y, f[N], ans[N];
vector<int> v[N];
int ksm(int x, int k) {
  int base = 1;
  while (k) {
    if (k & 1) base = 1ll * base * x % mod;
    k >>= 1;
    x = 1ll * x * x % mod;
  }
  return base;
}
int gcd(int x, int y) {
  if (!x) return y;
  return gcd(y % x, x);
}
void dfs(int u, int fa, int p) {
  int siz = 0;
  for (int i = 0; i < v[u].size(); i++) {
    if (v[u][i] == fa) continue;
    dfs(v[u][i], u, p);
    siz++;
  }
  if (!siz) {
    f[fa]++;
  } else {
    if (fa) {
      if (f[u] % p == 0)
        f[fa]++;
      else
        f[u]++;
    }
  }
}
int main() {
  T = read();
  while (T--) {
    n = read();
    for (int i = 1; i < n; i++) {
      x = read();
      y = read();
      v[x].push_back(y);
      v[y].push_back(x);
    }
    int tmp = ksm(2, n - 1);
    int pre = 0;
    for (int i = 2; i < n; i++) {
      if ((n - 1) % i == 0) {
        memset(f, 0, (n + 1) * 4);
        dfs(1, 0, i);
        int mi = 0;
        bool flag = 0;
        for (int k = 1; k <= n; k++) {
          mi = gcd(f[k], mi);
          if (f[k] % i) flag = 1;
        }
        if (!flag && mi == i) ans[i] = 1, pre++;
      }
    }
    cout << (tmp - pre + mod) % mod << " ";
    for (int i = 2; i <= n; i++) {
      cout << ans[i] << " ";
    }
    memset(ans, 0, (n + 1) * 4);
    for (int i = 1; i <= n; i++) v[i].clear();
    puts("");
  }
  return 0;
}

```