#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
void chkmin(T1 &x, T2 y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
void chkmax(T1 &x, T2 y) {
  if (x < y) x = y;
}
namespace fastio {
char rbuf[1 << 23], *p1 = rbuf, *p2 = rbuf, wbuf[1 << 23], *p3 = wbuf;
inline char getc() {
  return p1 == p2 &&
                 (p2 = (p1 = rbuf) + fread(rbuf, 1, 1 << 23, stdin), p1 == p2)
             ? -1
             : *p1++;
}
inline void putc(char x) { (*p3++ = x); }
template <typename T>
void read(T &x) {
  x = 0;
  char c = getchar();
  T neg = 0;
  while (!isdigit(c)) neg |= !(c ^ '-'), c = getchar();
  while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  if (neg) x = (~x) + 1;
}
template <typename T>
void recursive_print(T x) {
  return (!x) ? void() : (recursive_print(x / 10), putc(x % 10 ^ 48), void());
}
template <typename T>
void print(T x) {
  (!x) && (putc('0'), 0);
  (x < 0) && (putc('-'), x = ~x + 1);
  recursive_print(x);
}
template <typename T>
void print(T x, char c) {
  (!x) && (putc('0'), 0);
  (x < 0) && (putc('-'), x = ~x + 1);
  recursive_print(x);
  putc(c);
}
void print_final() { fwrite(wbuf, 1, p3 - wbuf, stdout); }
}  // namespace fastio
const int MOD = 998244353;
const int MAXN = 1e5;
int n, hd[MAXN + 5], to[MAXN * 2 + 5], nxt[MAXN * 2 + 5], ec = 0, deg[MAXN + 5],
                                                          pw2[MAXN + 5];
void adde(int u, int v) {
  to[++ec] = v;
  nxt[ec] = hd[u];
  hd[u] = ec;
}
int res[MAXN + 5], vis[MAXN + 5], cdeg[MAXN + 5], lft[MAXN + 5];
void solve() {
  scanf("%d", &n);
  ec = 0;
  for (int i = 1; i <= n; i++) hd[i] = deg[i] = res[i] = 0;
  int cnt = 0;
  for (int i = 1, u, v; i < n; i++)
    scanf("%d%d", &u, &v), deg[u]++, deg[v]++, adde(u, v), adde(v, u);
  for (int i = 2; i <= n - 1; i++) {
    if ((n - 1) % i) continue;
    queue<int> q;
    bool flg = 1;
    for (int j = 1; j <= n; j++) vis[j] = 0, cdeg[j] = lft[j] = deg[j];
    for (int j = 1; j <= n; j++)
      if (deg[j] == 1) q.push(j);
    while (!q.empty()) {
      int x = q.front();
      q.pop();
      vis[x] = 1;
      int y = 0;
      for (int e = hd[x]; e; e = nxt[e])
        if (!vis[to[e]]) y = to[e];
      if (!y) {
        if (cdeg[x] % i) {
          flg = 0;
          break;
        }
      } else {
        if (cdeg[x] % i == 0)
          cdeg[y]--;
        else if (cdeg[x] % i == 1)
          cdeg[x]--;
        else {
          flg = 0;
          break;
        }
        if ((--lft[y]) <= 1) q.push(y);
      }
    }
    res[i] = flg;
    cnt += flg;
  }
  res[1] = pw2[n - 1];
  for (int i = n; i; i--)
    for (int j = i << 1; j <= n; j += i) res[i] = (res[i] - res[j] + MOD) % MOD;
  for (int i = 1; i <= n; i++) printf("%d%c", res[i], " \n"[i == n]);
}
int main() {
  for (int i = (pw2[0] = 1); i <= MAXN; i++) pw2[i] = (pw2[i - 1] << 1) % MOD;
  int qu;
  scanf("%d", &qu);
  while (qu--) solve();
  return 0;
}

### Prompt

Your task is to create 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;
template <typename T1, typename T2>
void chkmin(T1 &x, T2 y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
void chkmax(T1 &x, T2 y) {
  if (x < y) x = y;
}
namespace fastio {
char rbuf[1 << 23], *p1 = rbuf, *p2 = rbuf, wbuf[1 << 23], *p3 = wbuf;
inline char getc() {
  return p1 == p2 &&
                 (p2 = (p1 = rbuf) + fread(rbuf, 1, 1 << 23, stdin), p1 == p2)
             ? -1
             : *p1++;
}
inline void putc(char x) { (*p3++ = x); }
template <typename T>
void read(T &x) {
  x = 0;
  char c = getchar();
  T neg = 0;
  while (!isdigit(c)) neg |= !(c ^ '-'), c = getchar();
  while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  if (neg) x = (~x) + 1;
}
template <typename T>
void recursive_print(T x) {
  return (!x) ? void() : (recursive_print(x / 10), putc(x % 10 ^ 48), void());
}
template <typename T>
void print(T x) {
  (!x) && (putc('0'), 0);
  (x < 0) && (putc('-'), x = ~x + 1);
  recursive_print(x);
}
template <typename T>
void print(T x, char c) {
  (!x) && (putc('0'), 0);
  (x < 0) && (putc('-'), x = ~x + 1);
  recursive_print(x);
  putc(c);
}
void print_final() { fwrite(wbuf, 1, p3 - wbuf, stdout); }
}  // namespace fastio
const int MOD = 998244353;
const int MAXN = 1e5;
int n, hd[MAXN + 5], to[MAXN * 2 + 5], nxt[MAXN * 2 + 5], ec = 0, deg[MAXN + 5],
                                                          pw2[MAXN + 5];
void adde(int u, int v) {
  to[++ec] = v;
  nxt[ec] = hd[u];
  hd[u] = ec;
}
int res[MAXN + 5], vis[MAXN + 5], cdeg[MAXN + 5], lft[MAXN + 5];
void solve() {
  scanf("%d", &n);
  ec = 0;
  for (int i = 1; i <= n; i++) hd[i] = deg[i] = res[i] = 0;
  int cnt = 0;
  for (int i = 1, u, v; i < n; i++)
    scanf("%d%d", &u, &v), deg[u]++, deg[v]++, adde(u, v), adde(v, u);
  for (int i = 2; i <= n - 1; i++) {
    if ((n - 1) % i) continue;
    queue<int> q;
    bool flg = 1;
    for (int j = 1; j <= n; j++) vis[j] = 0, cdeg[j] = lft[j] = deg[j];
    for (int j = 1; j <= n; j++)
      if (deg[j] == 1) q.push(j);
    while (!q.empty()) {
      int x = q.front();
      q.pop();
      vis[x] = 1;
      int y = 0;
      for (int e = hd[x]; e; e = nxt[e])
        if (!vis[to[e]]) y = to[e];
      if (!y) {
        if (cdeg[x] % i) {
          flg = 0;
          break;
        }
      } else {
        if (cdeg[x] % i == 0)
          cdeg[y]--;
        else if (cdeg[x] % i == 1)
          cdeg[x]--;
        else {
          flg = 0;
          break;
        }
        if ((--lft[y]) <= 1) q.push(y);
      }
    }
    res[i] = flg;
    cnt += flg;
  }
  res[1] = pw2[n - 1];
  for (int i = n; i; i--)
    for (int j = i << 1; j <= n; j += i) res[i] = (res[i] - res[j] + MOD) % MOD;
  for (int i = 1; i <= n; i++) printf("%d%c", res[i], " \n"[i == n]);
}
int main() {
  for (int i = (pw2[0] = 1); i <= MAXN; i++) pw2[i] = (pw2[i - 1] << 1) % MOD;
  int qu;
  scanf("%d", &qu);
  while (qu--) solve();
  return 0;
}

```