1554_E. You_128
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.
import sys

input = lambda: sys.stdin.readline().rstrip("\r\n")
MOD = 998244353

t = int(input())
while t > 0:
    t -= 1
    n = int(input())
    g = [[] for i in range(n)]
    for i in range(n - 1):
        x, y = map(int, input().split())
        g[x - 1] += [y - 1]
        g[y - 1] += [x - 1]
    f = [0] * n
    parent = [0] * n
    f[1] = pow(2, n - 1, MOD)
    order = [0]
    for v in order:
        for u in g[v]:
            if u != parent[v]:
                parent[u] = v
                order += [u]


    def dfs(k):
        size = [0] * n
        for v in reversed(order):
            if size[v] % k == 0:
                if v != 0:
                    size[parent[v]] += 1
            elif v == 0 or (size[v] + 1) % k != 0:
                return False
        return True


    for i in range(2, n):
        if (n - 1) % i == 0:
            f[i] = int(dfs(i))
    h = [0] * (n + 1)
    for i in range(n - 1, 0, -1):
        h[i] = f[i]
        for j in range(i * 2, n, i):
            h[i] -= h[j]
    print(*(x for x in h[1:n + 1]))
