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;
typedef long long LL;
typedef pair<int, int> P;
template<typename T> inline void chkmin(T &a, const T &b) { a = a < b ? a : b; }
template<typename T> inline void chkmax(T &a, const T &b) { a = a > b ? a : b; }

const int MAXN = 1 << 19, MOD = 924844033;
LL modpow(LL a, int b) {
	LL res = 1;
	for (; b; b >>= 1) {
		if (b & 1) res = res * a % MOD;
		a = a * a % MOD;
	}
	return res;
}

void ntt(LL *a, int n, int inv) {
	for (int i = 1, j = n >> 1; i < n; i++) {
		if (i < j) swap(a[i], a[j]);
		for (int k = n >> 1; (j ^= k) < k; k >>= 1);
	}
	for (int h = 2; h <= n; h <<= 1) {
		int hh = h >> 1, wn = modpow(5, inv ? MOD - 1 - (MOD - 1) / h : (MOD - 1) / h);
		for (int i = 0; i < n; i += h) {
			LL w = 1;
			for (int j = i; j < i + hh; j++) {
				LL x = a[j], y = a[j + hh] * w % MOD;
				a[j] = (x + y) % MOD;
				a[j + hh] = (x - y + MOD) % MOD;
				w = w * wn % MOD;
			}
		}
	}
	if (inv) {
		LL mul = modpow(n, MOD - 2);
		for (int i = 0; i < n; i++) a[i] = a[i] * mul % MOD;
	}
}

LL A[MAXN], B[MAXN], fac[MAXN], inv[MAXN];
int cnt[MAXN], sz[MAXN], n;
vector<int> gra[MAXN];

void dfs(int u, int fa) {
	sz[u] = 1;
	for (int v : gra[u]) if (v != fa) {
		dfs(v, u);
		sz[u] += sz[v];
		++cnt[sz[v]];
	}
	++cnt[n - sz[u]];
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i < n; i++) {
		int u, v; scanf("%d%d", &u, &v);
		gra[u].push_back(v), gra[v].push_back(u);
	}
	for (int i = fac[0] = 1; i <= n; i++) fac[i] = fac[i - 1] * i % MOD;
	inv[n] = modpow(fac[n], MOD - 2);
	for (int i = n; i > 0; i--) inv[i - 1] = inv[i] * i % MOD;
	dfs(1, 0);
	for (int i = 0; i < n; i++) {
		A[i] = (LL)cnt[n - i] * fac[n - i] % MOD;
		B[i] = inv[i];
	}
	int m = 1;
	for (; m < n + n - 1; m <<= 1);
	ntt(A, m, 0), ntt(B, m, 0);
	for (int i = 0; i < m; i++) A[i] = A[i] * B[i] % MOD;
	ntt(A, m, 1);
	for (int i = 1; i <= n; i++)
		printf("%lld\n", (n * fac[n] % MOD * inv[n - i] - A[n - i] + MOD) % MOD * inv[i] % MOD);
	return 0;
}