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
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;

#define iter(i, n) forw(i, 1, n)
#define iter0(i, n) for (int i = 0; i < n; ++i)
#define forw(i, a, b) for (int i = a; i <= b; ++i)

const int g = 5;
const int mod = 924844033;

#define NR 211000
#define MR 811000

vector<int> e[NR];
int n, a[MR], b[MR], sz[NR], fac[NR], inv[NR], N;

int pr(int a, int z) {
	int s = 1;
	while (z > 0) {
		if (z % 2 == 1) s = 1ll * s * a % mod;
		a = 1ll * a * a % mod;
		z /= 2;
	}
	return s;
}

void fft(int *a, int n, int ty) {
	for (int i = n >> 1, j = 1, k; j < n - 1; ++j) {
		if (i > j) swap(a[i], a[j]);
		for (k = n >> 1; k <= i; k >>= 1) i ^= k;
		i ^= k;
	}

	for (int m = 2; m <= n; m <<= 1) {
		int h = m >> 1, wm = pr(g, (mod - 1) / m * (ty == +1 ? 1 : (m - 1)));

		for (int i = 0; i < n; i += m) {
			int w = 1;
			for (int j = i; j < i + h; ++j) {
				int u = a[j], v = 1ll * w * a[j + h] % mod;
				a[j] = (u + v) % mod;
				a[j + h] = (u - v + mod) % mod;
				w = 1ll * w * wm % mod;
			}
		}
	}

	if (ty == -1) {
		int iv = pr(n, mod - 2);
		iter0(i, n) a[i] = 1ll * a[i] * iv % mod;
	}
}

void dfs(int x, int fa) {
	sz[x] = 1;
	for (int v : e[x]) if (v != fa) {
		dfs(v, x), sz[x] += sz[v]; ++a[sz[v]];
	}
	++a[n - sz[x]];
}

int binom(int n, int k) { return 1ll * fac[n] * inv[k] % mod * inv[n - k] % mod; }

int main() {
	scanf("%d", &n);
	iter(i, n - 1) {
		int x, y;
		scanf("%d%d", &x, &y);
		e[x].push_back(y), e[y].push_back(x);
	}

	dfs(1, 0);


	for (N = 1; N < 2 * n; N <<= 1);

	fac[0] = inv[0] = 1;
	a[0] = 0;
	iter(i, n) {
		fac[i] = 1ll * fac[i - 1] * i % mod;
		
		inv[i] = pr(fac[i], mod - 2);
	}
	

	


	iter(i, n) a[i] = 1ll * fac[i] * a[i] % mod;
	forw(i, 0, n - 1) {
		b[n - i] = inv[i];
	}

	//printf("%d\n", a[2*n]);

	fft(a, N, 1), fft(b, N, 1);
	iter0(i, N) a[i] = 1ll * a[i] * b[i] % mod;
	fft(a, N, -1);

	iter(i, n) printf("%lld\n", (1ll * n * binom(n, i) % mod + mod - 1ll * a[n + i] * inv[i] % mod) % mod);
	return 0;
}