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>
using namespace std;

const int _ = 1 << 19 | 1 , MOD = 924844033 , G = 5;
struct Edge{int end , upEd;}Ed[_ << 1];
int arr[_] , tp[_] , N , head[_] , cntEd , jc[_] , inv[_];
void addEd(int a , int b){Ed[++cntEd] = (Edge){b , head[a]}; head[a] = cntEd;}
int upd(int x){return x + (x >> 31 & MOD);}

int poww(long long a , int b){
	int times = 1;
	while(b){
		if(b & 1) times = times * a % MOD;
		a = a * a % MOD; b >>= 1;
	}
	return times;
}

int dfs(int x , int p){
	int sz = 1;
	for(int i = head[x] ; i ; i = Ed[i].upEd)
		if(Ed[i].end != p){int t = dfs(Ed[i].end , x); ++arr[t]; ++arr[N - t]; sz += t;}
	return sz;
}

int dir[_] , need , invnd;
void init(int len){
	need = 1; while(need < len) need <<= 1;
	invnd = poww(need , MOD - 2);
	for(int i = 1 ; i < need ; ++i)
		dir[i] = (dir[i >> 1] >> 1) | (i & 1 ? need >> 1 : 0);
}

void DFT(int *arr){
	for(int i = 1 ; i < need ; ++i)
		if(i < dir[i]) swap(arr[i] , arr[dir[i]]);
	for(int i = 1 ; i < need ; i <<= 1){
		int wn = poww(G , MOD / i / 2);
		for(int j = 0 ; j < need ; j += i << 1){
			long long w = 1;
			for(int k = 0 ; k < i ; ++k , w = w * wn % MOD){
				int x = arr[j + k] , y = arr[i + j + k] * w % MOD;
				arr[j + k] = upd(x + y - MOD); arr[i + j + k] = upd(x - y);
			}
		}
	}
}

void IDFT(int *arr){
	reverse(arr + 1 , arr + need); DFT(arr);
	for(int i = 0 ; i < need ; ++i) arr[i] = 1ll * arr[i] * invnd % MOD;
}

int binom(int a , int b){return 1ll * jc[a] * inv[b] % MOD * inv[a - b] % MOD;}

int main(){
	ios::sync_with_stdio(0); cin >> N;
	for(int i = 1 , x , y ; i < N ; ++i){cin >> x >> y; addEd(x , y); addEd(y , x);}
	dfs(1 , 0); jc[0] = 1;
	for(int i = 1 ; i <= N ; ++i) jc[i] = 1ll * jc[i - 1] * i % MOD;
	for(int i = 1 ; i <= N ; ++i) arr[i] = 1ll * arr[i] * jc[i] % MOD;
	inv[N] = poww(jc[N] , MOD - 2);
	for(int i = N - 1 ; i >= 0 ; --i) inv[i] = inv[i + 1] * (i + 1ll) % MOD;
	memcpy(tp , inv , sizeof(int) * (N + 1)); reverse(tp , tp + N + 1);
	init(2 * N + 1); DFT(arr); DFT(tp);
	for(int i = 0 ; i < need ; ++i) arr[i] = 1ll * arr[i] * tp[i] % MOD;
	IDFT(arr);
	for(int i = 1 ; i <= N ; ++i)
		cout << (1ll * N * binom(N , i) % MOD - 1ll * arr[N + i] * inv[i] % MOD + MOD) % MOD << endl;
	return 0;
}