Problem: One day, Takahashi was given the following problem from Aoki:

* You are given a tree with N vertices and an integer K. The vertices are numbered 1 through N. The edges are represented by pairs of integers (a_i, b_i).
* For a set S of vertices in the tree, let f(S) be the minimum number of the vertices in a subtree of the given tree that contains all vertices in S.
* There are <image> ways to choose K vertices from the trees. For each of them, let S be the set of the chosen vertices, and find the sum of f(S) over all <image> ways.
* Since the answer may be extremely large, print it modulo 924844033(prime).



Since it was too easy for him, he decided to solve this problem for all K = 1,2,...,N.

Constraints

* 2 ≦ N ≦ 200,000
* 1 ≦ a_i, b_i ≦ N
* The given graph is a tree.

Input

The input is given from Standard Input in the following format:


N
a_1 b_1
a_2 b_2
:
a_{N-1} b_{N-1}


Output

Print N lines. The i-th line should contain the answer to the problem where K=i, modulo 924844033.

Examples

Input

3
1 2
2 3


Output

3
7
3


Input

4
1 2
1 3
1 4


Output

4
15
13
4


Input

7
1 2
2 3
2 4
4 5
4 6
6 7


Output

7
67
150
179
122
45
7

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

const int Maxn = 1 << 19 | 5, p = 924844033;
int cnt, rev[Maxn],  siz[Maxn], head[Maxn];
long long fac[Maxn], inv[Maxn];
void get_rev(int n)
{
	for (int i = 0; i < n; i++)
		rev[i] = (rev[i >> 1] >> 1) | ((i & 1) * (n >> 1));
}
long long fast_pow(long long x, long long y)
{
	long long ans = 1, now = x;
	while (y)
	{
		if (y & 1) ans = ans * now % p;
		now = now * now % p;
		y >>= 1;
	}
	return ans;
}
void NTT(long long now[], int n, bool type = false)
{
	get_rev(n);
	for (int i = 0; i < n; i++)
		if (i < rev[i]) swap(now[i], now[rev[i]]);
	for (int i = 1; i < n; i <<= 1)
	{
		long long w = fast_pow(type ? 554906420 : 5, (p - 1) / (i << 1));
		for (int j = 0; j < n; j += (i << 1))
		{
			long long l = 1;
			for (int k = j; k < i + j; k++, l = l * w % p)
			{
				long long x = now[i + k] * l % p, y = now[k];
				now[k] = (x + y) % p;
				now[i + k] = (y - x + p) % p;
			}
		}
	}
	if (type)
	{
		long long inv = fast_pow(n, p - 2);
		for (int i = 0; i < n; i++)
			now[i] = now[i] * inv % p;
	}
}
int n;
void init(void)
{
	fac[0] = 1;
	for (int i = 1; i <= n; i++)
		fac[i] = fac[i - 1] * i % p;
	inv[n] = fast_pow(fac[n], p - 2);
	for (int i = n - 1; i >= 0; i--)
		inv[i] = inv[i + 1] * (i + 1) % p;
}
struct edg
{
	int nxt, to;
} edge[2 * Maxn];
void add(int x, int y)
{
	edge[++cnt] = (edg){head[x], y};
	head[x] = cnt;
}
void dfs(int u, int fa)
{
	siz[u] = 1;
	for (int i = head[u]; i; i = edge[i].nxt)
	{
		int to = edge[i].to;
		if (to != fa)
			dfs(to, u), siz[u] += siz[to];
	}
}
int lower(int x)
{
	int tmp = 1;
	for (; tmp <= x; tmp <<= 1);
	return tmp;
}
long long C(int x, int y)
{
	return fac[x] * inv[y] % p * inv[x - y] % p;
}
long long A[Maxn], B[Maxn];
int main()
{
	scanf("%d", &n);
	for (int i = 1; i < n; i++)
	{
		int x, y;
		scanf("%d%d", &x, &y);
		add(x, y);
		add(y, x);
	}
	dfs(1, 0);
	init();
	for (int i = 2; i <= n; i++)
	{
		A[n - siz[i]] += fac[n - siz[i]];
		A[siz[i]] += fac[siz[i]];
	}
	int len = lower(n) << 1;
	for (int i = 0; i <= n; i++)
		B[n - i] = inv[i];
	NTT(A, len), NTT(B, len);
	for (int i = 0; i < len; i++)
		(A[i] *= B[i]) %= p;
	NTT(A, len, true);
	for (int i = n + 1; i <= 2 * n; i++)
		printf("%lld\n", (n * C(n, i - n) % p - A[i] * inv[i - n] % p + p) % p);
	return 0;
}