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 <iostream>
#include <vector>

using namespace std;

constexpr long long MOD = 924844033;
constexpr long long g = 5;

vector<int> edges[200001];

int n;
long long a[524288], b[524288], om[524288], invom[524288];
long long inv[200001], invfac[200001];

long long qpow(long long x, long long y) {
	long long result = 1;
	while (y != 0) {
		if (y % 2 == 1) result = result * x % MOD;
		x = x * x % MOD;
		y /= 2;
	}
	return result;
}

void fft(long long *a, int k, long long *om) {
	for (int i = 0, j = 0; i < k; ++i) {
		if (i < j) swap(a[i], a[j]);
		for (int l = k / 2; (j ^= l) < l; l /= 2);
	}

	for (int i = 2; i <= k; i *= 2) {
		for (int j = 0; j < k; j += i) {
			for (int l = 0; l < i / 2; ++l) {
				long long t = a[j + i / 2 + l] * om[k / i * l] % MOD;
				a[j + i / 2 + l] = (a[j + l] + MOD - t) % MOD;
				a[j + l] = (a[j + l] + t) % MOD;
			}
		}
	}
}

int dfs(int x, int p) {
	int size = 1;
	for (auto y: edges[x]) if (y != p) {
		size += dfs(y, x);
	}
	
	if (p != 0) {
		--a[size];
		--a[n - size];
	}
	return size;
}

int main() {
#ifdef ONLINE_JUDGE
	ios_base::sync_with_stdio(false);
#endif

	cin >> n;
	for (int i = 1; i < n; ++i) {
		int a, b;
		cin >> a >> b;
		edges[a].push_back(b);
		edges[b].push_back(a);
	}

	dfs(1, 0);
	a[n] = n;

	inv[1] = 1;
	for (int i = 2; i <= n; ++i) inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;

	long long fac = 1;
	b[n] = invfac[0] = 1;
	for (int i = 1; i <= n; ++i) {
		fac = fac * i % MOD;
		a[i] = a[i] * fac % MOD;
		b[n - i] = invfac[i] = invfac[i - 1] * inv[i] % MOD;
	}

	int k = 1;
	while (k < n * 2 + 1) k *= 2;

	long long r = qpow(g, (MOD - 1) / k);
	om[0] = invom[0] = 1;
	for (int i = 1; i < k; ++i) om[i] = invom[k - i] = om[i - 1] * r % MOD;

	fft(a, k, om);
	fft(b, k, om);
	for (int i = 0; i < k; ++i) a[i] = a[i] * b[i] % MOD;
	fft(a, k, invom);

	k = qpow(k, MOD - 2);
	for (int i = 1; i <= n; ++i)
		cout << a[i + n] * k % MOD * invfac[i] % MOD << '\n';
	
	return 0;
}