#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int n = 0, f = 1, ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    n = n * 10 + ch - '0';
    ch = getchar();
  }
  return n * f;
}
char str[200005];
int a[200005];
vector<int> v[200005];
int k;
int dp[200005][2], rd[200005];
int da[200005], jda[200005];
void dfs(int x, int f) {
  if (rd[x] < k) {
    dp[x][0] = 1;
    for (int i = 0; i < v[x].size(); i++) {
      if (v[x][i] == f) continue;
      dfs(v[x][i], x);
      dp[x][0] = 1LL * dp[x][0] * dp[v[x][i]][1] % 998244353;
    }
    dp[x][1] = 0;
    return;
  }
  int cj = 1, sl = 0;
  for (int i = 0; i < v[x].size(); i++) {
    if (v[x][i] == f) continue;
    dfs(v[x][i], x);
    if (dp[v[x][i]][1] == 0) {
      cj = 1LL * dp[v[x][i]][0] * cj % 998244353;
      sl++;
    }
  }
  for (int i = 0; i < k; i++) da[i] = 0;
  da[sl % k] = cj;
  for (int i = 0; i < v[x].size(); i++) {
    if (v[x][i] == f) continue;
    if (dp[v[x][i]][1] == 0) continue;
    for (int j = 0; j < k; j++) jda[j] = da[j];
    for (int j = 0; j < k; j++) {
      da[j] = 1LL * jda[j] * dp[v[x][i]][1] % 998244353;
      if (j == 0)
        da[j] =
            (da[j] + 1LL * jda[k - 1] * dp[v[x][i]][0] % 998244353) % 998244353;
      else
        da[j] =
            (da[j] + 1LL * jda[j - 1] * dp[v[x][i]][0] % 998244353) % 998244353;
    }
  }
  dp[x][0] = da[0];
  dp[x][1] = da[k - 1];
}
int tans[200005];
int main() {
  int t, n, x, y;
  t = read();
  for (int greg = 1; greg <= t; greg++) {
    n = read();
    for (int i = 1; i <= n - 1; i++) {
      x = read();
      y = read();
      v[x].push_back(y);
      v[y].push_back(x);
      rd[x]++;
      rd[y]++;
    }
    for (int i = 1; i <= n; i++) {
      if ((n - 1) % i == 0) {
        k = i;
        dfs(1, 0);
        tans[i] = dp[1][0];
      } else
        tans[i] = 0;
    }
    for (int i = n; i >= 1; i--) {
      for (int j = 2 * i; j <= n; j += i)
        tans[i] = (tans[i] + 998244353 - tans[j]) % 998244353;
    }
    for (int i = 1; i <= n; i++) printf("%d ", tans[i]);
    printf("\n");
    for (int i = 1; i <= n; i++) {
      v[i].clear();
      dp[i][0] = dp[i][1] = 0;
      rd[i] = 0;
    }
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 int read() {
  int n = 0, f = 1, ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    n = n * 10 + ch - '0';
    ch = getchar();
  }
  return n * f;
}
char str[200005];
int a[200005];
vector<int> v[200005];
int k;
int dp[200005][2], rd[200005];
int da[200005], jda[200005];
void dfs(int x, int f) {
  if (rd[x] < k) {
    dp[x][0] = 1;
    for (int i = 0; i < v[x].size(); i++) {
      if (v[x][i] == f) continue;
      dfs(v[x][i], x);
      dp[x][0] = 1LL * dp[x][0] * dp[v[x][i]][1] % 998244353;
    }
    dp[x][1] = 0;
    return;
  }
  int cj = 1, sl = 0;
  for (int i = 0; i < v[x].size(); i++) {
    if (v[x][i] == f) continue;
    dfs(v[x][i], x);
    if (dp[v[x][i]][1] == 0) {
      cj = 1LL * dp[v[x][i]][0] * cj % 998244353;
      sl++;
    }
  }
  for (int i = 0; i < k; i++) da[i] = 0;
  da[sl % k] = cj;
  for (int i = 0; i < v[x].size(); i++) {
    if (v[x][i] == f) continue;
    if (dp[v[x][i]][1] == 0) continue;
    for (int j = 0; j < k; j++) jda[j] = da[j];
    for (int j = 0; j < k; j++) {
      da[j] = 1LL * jda[j] * dp[v[x][i]][1] % 998244353;
      if (j == 0)
        da[j] =
            (da[j] + 1LL * jda[k - 1] * dp[v[x][i]][0] % 998244353) % 998244353;
      else
        da[j] =
            (da[j] + 1LL * jda[j - 1] * dp[v[x][i]][0] % 998244353) % 998244353;
    }
  }
  dp[x][0] = da[0];
  dp[x][1] = da[k - 1];
}
int tans[200005];
int main() {
  int t, n, x, y;
  t = read();
  for (int greg = 1; greg <= t; greg++) {
    n = read();
    for (int i = 1; i <= n - 1; i++) {
      x = read();
      y = read();
      v[x].push_back(y);
      v[y].push_back(x);
      rd[x]++;
      rd[y]++;
    }
    for (int i = 1; i <= n; i++) {
      if ((n - 1) % i == 0) {
        k = i;
        dfs(1, 0);
        tans[i] = dp[1][0];
      } else
        tans[i] = 0;
    }
    for (int i = n; i >= 1; i--) {
      for (int j = 2 * i; j <= n; j += i)
        tans[i] = (tans[i] + 998244353 - tans[j]) % 998244353;
    }
    for (int i = 1; i <= n; i++) printf("%d ", tans[i]);
    printf("\n");
    for (int i = 1; i <= n; i++) {
      v[i].clear();
      dp[i][0] = dp[i][1] = 0;
      rd[i] = 0;
    }
  }
  return 0;
}

```