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
// Many Easy Problems
// * frank_c1
// * 2017 / 09 / 29
 
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;

LL pow_mod(LL b, LL p, LL k) {
	LL ret = 1;
	for (; p; p >>= 1) {
		if (p & 1) ret = ret * b % k;
		b = b * b % k;
	} return ret;
}

const int maxn = (int)(1e6) + 5;
const int mo = 924844033;
const int gi = 5;
int cnt[maxn], siz[maxn];
LL f[maxn], g[maxn], fac[maxn], ivf[maxn];
vector<int> nxt[maxn];
int n;

inline LL C(int n, int m) {
	if (m < 0 || n - m < 0) return 0;
	return (fac[n] * ivf[m] % mo) * ivf[n - m] % mo;
}

void dfs(int x, int f) {
	siz[x] = 1;
	for (int i = 0; i < (int)nxt[x].size(); ++i) if (nxt[x][i] != f) {
		dfs(nxt[x][i], x); siz[x] += siz[nxt[x][i]];
	} if (f) {
		++cnt[siz[x]]; ++cnt[n - siz[x]];
	}
}

void NTT(LL a[], int n, int d = 1) {
	for (int i = (n >> 1), j = 1; j < n; ++j) {
		if (i < j) swap(a[i], a[j]);
		int k; for (k = (n >> 1); i & k; i ^= k, k >>= 1);
		i ^= k;
	}
	for (int m = 2; m <= n; m <<= 1) {
		LL w = pow_mod(gi, (d * ((mo - 1) / m) + mo - 1) % (mo - 1), mo);
		for (int i = 0; i < n; i += m) {
			LL s = 1;
			for (int j = i; j < i + (m >> 1); ++j) {
				LL u = a[j], v = a[j + (m >> 1)] * s % mo;
				a[j] = (u + v) % mo; a[j + (m >> 1)] = (u - v + mo) % mo;
				s = s * w % mo;
			}
		}
	} 
	if (d == -1) {
		LL xinv = pow_mod(n, mo - 2, mo);
		for (int i = 0; i < n; ++i) a[i] = a[i] * xinv % mo;
	}
}

int main() {
	scanf("%d", &n);
	fac[0] = ivf[0] = 1;
	for (int i = 1; i <= n; ++i) fac[i] = fac[i - 1] * i % mo;
	ivf[n] = pow_mod(fac[n], mo - 2, mo);
	for (int i = n - 1; i >= 1; --i) ivf[i] = ivf[i + 1] * (i + 1) % mo;
	for (int i = 1; i < n; ++i) {
		int u, v; scanf("%d%d", &u, &v);
		nxt[u].push_back(v); nxt[v].push_back(u);
	} dfs(1, 0);
	int len = 0;
	while ((1 << len) <= (n << 1)) ++len;
	for (int i = 0; i < n; ++i) f[i] = ivf[i];
	for (int i = 1; i <= n; ++i) g[n - i + 1] = fac[i] * cnt[i] % mo;
	NTT(f, 1 << len); NTT(g, 1 << len);
	for (int i = 0; i < (1 << len); ++i) f[i] = f[i] * g[i] % mo;
	NTT(f, 1 << len, -1);
	for (int i = 1; i <= n; ++i) printf("%lld\n", (n * C(n, i) + f[n - i + 1] * (mo - ivf[i])) % mo);
 	return 0;
}