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 <cstdio>

inline void swap(int &x, int &y) { x ^= y ^= x ^= y; }

typedef long long LL;
const int MN = 200005;
const int MS = 524288;
const int Mod = 924844033;
const int G = 5, iG = 554906420;

inline int qPow(int b, int e) {
	int a = 1;
	for (; e; b = (LL)b * b % Mod, e >>= 1)
		if (e & 1) a = (LL)a * b % Mod;
	return a;
}

int N;
int h[MN], nxt[MN * 2], to[MN * 2], tot;
inline void ins(int x, int y) {
	nxt[++tot] = h[x], to[tot] = y, h[x] = tot;
}

int Inv[MN], Fac[MN], iFac[MN];
int R[MS], A[MS], B[MS];

void Init(int N) {
	Inv[1] = 1;
	for (int i = 2; i <= N; ++i) {
		Inv[i] = (LL)(Mod - Mod / i) * Inv[Mod % i] % Mod;
	}
	Fac[0] = iFac[0] = 1;
	for (int i = 1; i <= N; ++i) {
		Fac[i] = (LL)Fac[i - 1] * i % Mod;
		iFac[i] = (LL)iFac[i - 1] * Inv[i] % Mod;
	}
}

int DFS(int u, int f) {
	int sz = 1, sv;
	for (int i = h[u]; i; i = nxt[i]) if (to[i] != f) {
		sz += sv = DFS(to[i], u);
		--A[sv], --A[N - sv];
	} return sz;
}

void FNTT(int *A, int Sz, int Ty) {
	for (int i = 0; i < Sz; ++i)
		if (R[i] < i) swap(A[R[i]], A[i]);
	for (int j = 1; j < Sz; j <<= 1) {
		int j2 = j << 1, wn = qPow(~Ty ? G : iG, (Mod - 1) / j2);
		for (int i = 0; i < Sz; i += j2) {
			for (int k = 0, w = 1; k < j; ++k, w = (LL)w * wn % Mod) {
				int X = A[i + k], Y = (LL)w * A[i + j + k] % Mod;
				A[i + k] = X + Y; if (A[i + k] >= Mod) A[i + k] -= Mod;
				A[i + j + k] = X - Y; if (A[i + j + k] < 0) A[i + j + k] += Mod;
			}
		}
	}
	if (Ty == -1) {
		int InvSz = qPow(Sz, Mod - 2);
		for (int i = 0; i < Sz; ++i) A[i] = (LL)A[i] * InvSz % Mod;
	}
}

int main() {
	scanf("%d", &N);
	Init(N);
	for (int i = 1, x, y; i < N; ++i) {
		scanf("%d%d", &x, &y);
		ins(x, y), ins(y, x);
	}
	DFS(1, 0), A[N] = N;
	for (int i = 1; i <= N; ++i) A[i] = ((LL)A[i] * Fac[i] % Mod + Mod) % Mod;
	for (int i = 0; i < N; ++i) B[i] = iFac[N - 1 - i];
	int Sz = 1, Bt = 0;
	for (; Sz < N + N; Sz <<= 1, ++Bt) ;
	for (int i = 0; i < Sz; ++i) R[i] = R[i >> 1] >> 1 | (i & 1) << (Bt - 1);
	FNTT(A, Sz, 1), FNTT(B, Sz, 1);
	for (int i = 0; i < Sz; ++i) A[i] = (LL)A[i] * B[i] % Mod;
	FNTT(A, Sz, -1);
	for (int i = 1; i <= N; ++i) printf("%d\n", (LL)A[i + N - 1] * iFac[i] % Mod);
	return 0;
}