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 <bits/stdc++.h>

typedef long long ll;
typedef long long llong;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;

const ll MOD = 924844033;
ll pw(ll a, ll b) {
	ll ans = 1; while (b) {
		while (!(b & 1)) b >>= 1, a = (a * a) % MOD;
		ans = (ans * a) % MOD, --b;
	} return ans;
}
const ll B = 3597;
const int MAXN = 220000;

int was[MAXN];
int c[MAXN];
ll a[1 << 19];
ll b[1 << 19];
ll fc[MAXN];
ll bfc[MAXN];
vector<int> eds[MAXN];
int sz[MAXN];
int n;

void dfs1(int v) {
	sz[v] = 1;
	was[v] = 1;
	for (int u: eds[v]) {
		if (was[u])
			continue;
		dfs1(u);
		sz[v] += sz[u];
		++c[sz[u]];
	}
	if (sz[v] != n)
		++c[n - sz[v]];
}

int rev(int x, int k) {
	int ans = 0;
	for (int i = 0; i < k; ++i) {
		ans = ans * 2 + (x & 1);
		x >>= 1;
	}
	return ans;
}

void fft(ll *a, int k, int inv) {
	int n = (1 << k);
	for (int i = 0; i < n; ++i) {
		int x = rev(i, k);
		if (x > i)
			swap(a[x], a[i]);
	}
	for (int bl = 1; bl < n; bl *= 2) {
		ll wadd = pw(B, (1 << 20) / bl);
		if (inv)
			wadd = pw(wadd, MOD - 2);
		for (int i = 0; i < n; i += 2 * bl) {
			ll w = 1;
			for (int j = i; j < i + bl; ++j, w = (w * wadd) % MOD) {
				ll u = a[j];
				ll v = a[j + bl] * w % MOD;
				a[j] = (u + v) % MOD;
				a[j + bl] = (u - v + MOD) % MOD;
			}
		}
	}
	if (inv) {
		ll n2 = pw(n, MOD - 2);
		for (int j = 0; j < n; ++j)
			a[j] = (a[j] * n2) % MOD;
	}
}

int main() {
	scanf("%d", &n);
	fc[0] = bfc[0] = 1;
	for (int i = 1; i <= n; ++i)
		fc[i] = (fc[i - 1] * i) % MOD, bfc[i] = pw(fc[i], MOD - 2);
	for (int i = 0; i < n - 1; ++i) {
		int a, b;
		scanf("%d%d", &a, &b);
		--a, --b;
		eds[a].push_back(b);
		eds[b].push_back(a);
	}
	dfs1(0);
	for (int i = 0; i <= n; ++i)
		a[i] = (c[i] * fc[i]) % MOD;
	for (int i = 0; i <= n; ++i)
		b[i] = bfc[n - i];
	int k = 0;
	while ((1 << k) <= 2 * n)
		++k;
	fft(a, k, 0);
	fft(b, k, 0);
	for (int i = 0; i < (1 << k); ++i)
		a[i] = (a[i] * b[i]) % MOD;
	fft(a, k, 1);
	for (int i = 1; i <= n; ++i) {
		a[i] = a[i + n] * bfc[i] % MOD;
		ll go = fc[n] * bfc[i] % MOD * bfc[n - i] % MOD * n % MOD;
		go = (go - a[i] + MOD) % MOD;
		printf("%lld\n", go);
	}
	return 0;
}


