#include <bits/stdc++.h>
const int MAXN = 1e5, MOD = 998244353;
int n, ecnt, head[MAXN + 5], f[MAXN + 5];
struct Edge {
  int to, nxt;
} graph[MAXN * 2 + 5];
inline void subeq(int& a, const int b) { (a -= b) < 0 && (a += MOD); }
inline int mul(const int a, const int b) { return int(1ll * a * b % MOD); }
inline int mpow(int a, int b) {
  int ret = 1;
  for (; b; a = mul(a, a), b >>= 1) ret = mul(ret, b & 1 ? a : 1);
  return ret;
}
inline void link(const int u, const int v) {
  graph[++ecnt] = {v, head[u]}, head[u] = ecnt;
  graph[++ecnt] = {u, head[v]}, head[v] = ecnt;
}
inline bool check(const int u, const int fa, const int k) {
  static int val[MAXN + 5];
  val[u] = 0;
  for (int i = head[u], v; i; i = graph[i].nxt) {
    if ((v = graph[i].to) != fa) {
      if (!check(v, u, k)) return false;
    }
  }
  if (val[u] % k) {
    if (!fa || (val[u] + 1) % k) return false;
  } else if (fa)
    ++val[fa];
  return true;
}
inline void allClear() {
  ecnt = 0;
  for (int i = 1, repi = n; i <= repi; ++i) head[i] = 0;
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n), allClear();
    for (int i = 2, repi = n; i <= repi; ++i) {
      int u, v;
      scanf("%d %d", &u, &v);
      link(u, v);
    }
    f[1] = mpow(2, n - 1);
    for (int i = 2, repi = n - 1; i <= repi; ++i)
      f[i] = !((n - 1) % i) && check(1, 0, i);
    f[n] = 0;
    for (int i = n, peri = 1; i >= peri; --i)
      for (int j = 2, repj = n / i; j <= repj; ++j) subeq(f[i], f[i * j]);
    for (int i = 1, repi = n; i <= repi; ++i)
      printf("%d%c", f[i], i < n ? ' ' : '\n');
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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>
const int MAXN = 1e5, MOD = 998244353;
int n, ecnt, head[MAXN + 5], f[MAXN + 5];
struct Edge {
  int to, nxt;
} graph[MAXN * 2 + 5];
inline void subeq(int& a, const int b) { (a -= b) < 0 && (a += MOD); }
inline int mul(const int a, const int b) { return int(1ll * a * b % MOD); }
inline int mpow(int a, int b) {
  int ret = 1;
  for (; b; a = mul(a, a), b >>= 1) ret = mul(ret, b & 1 ? a : 1);
  return ret;
}
inline void link(const int u, const int v) {
  graph[++ecnt] = {v, head[u]}, head[u] = ecnt;
  graph[++ecnt] = {u, head[v]}, head[v] = ecnt;
}
inline bool check(const int u, const int fa, const int k) {
  static int val[MAXN + 5];
  val[u] = 0;
  for (int i = head[u], v; i; i = graph[i].nxt) {
    if ((v = graph[i].to) != fa) {
      if (!check(v, u, k)) return false;
    }
  }
  if (val[u] % k) {
    if (!fa || (val[u] + 1) % k) return false;
  } else if (fa)
    ++val[fa];
  return true;
}
inline void allClear() {
  ecnt = 0;
  for (int i = 1, repi = n; i <= repi; ++i) head[i] = 0;
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n), allClear();
    for (int i = 2, repi = n; i <= repi; ++i) {
      int u, v;
      scanf("%d %d", &u, &v);
      link(u, v);
    }
    f[1] = mpow(2, n - 1);
    for (int i = 2, repi = n - 1; i <= repi; ++i)
      f[i] = !((n - 1) % i) && check(1, 0, i);
    f[n] = 0;
    for (int i = n, peri = 1; i >= peri; --i)
      for (int j = 2, repj = n / i; j <= repj; ++j) subeq(f[i], f[i * j]);
    for (int i = 1, repi = n; i <= repi; ++i)
      printf("%d%c", f[i], i < n ? ' ' : '\n');
  }
  return 0;
}

```