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>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstring>
#include <vector>

#define For(i, j, k) for(int i = j; i <= k; i++)
#define Forr(i, j, k) for(int i = j; i >= k; i--)
#define pb push_back

using namespace std;

const int N = 1010010;
const int Mod = 924844033;
const int r = 5;

typedef long long LL;

LL Pow(LL x, LL e){
	LL ret = 1;
	while(e){
		if(e & 1) ret = ret * x % Mod;
		x = x * x % Mod;
		e >>= 1;
	}
	return ret;
}

int rev[N];

void DFT(LL* A, int n, bool inv){
	For(i, 0, n - 1) if(i < rev[i]) swap(A[i], A[rev[i]]);
	for(int i = 2; i <= n; i <<= 1){
		LL w = Pow(r, (Mod - 1) / i);
		if(inv) w = Pow(w, Mod - 2);
		for(int j = 0; j < n; j += i){
			LL cur = 1;
			for(int k = 0; k < (i >> 1); ++k){
				LL x = A[j + k], y = A[j + k + (i >> 1)] * cur % Mod;
				A[j + k] = (x + y) % Mod, A[j + k + (i >> 1)] = (x - y + Mod) % Mod;
				cur = cur * w % Mod;
			}
		}
	}
}

void NTT(LL *A, LL *B, int m){
	int n = 1;
	while(n <= m + 2) n <<= 1;
	For(i, 1, n - 1) rev[i] = (rev[i >> 1] >> 1) | (i & 1 ? n >> 1 : 0);
	DFT(A, n, 0), DFT(B, n, 0);
	For(i, 0, n - 1) A[i] = A[i] * B[i] % Mod;
	DFT(A, n, 1);
	LL inv = Pow(n, Mod - 2);
	For(i, 0, n - 1) A[i] = A[i] * inv % Mod;
}

LL coe[N], F[N], fac[N], rfac[N];
int n, sz[N];
vector<int> G[N];

void DFS(int o, int f){
	sz[o] = 1;
	for(int v : G[o]){
		if(v == f) continue;
		DFS(v, o);
		sz[o] += sz[v];
		coe[sz[v]]++;
	}
	if(n > sz[o]) coe[n - sz[o]]++;
}

LL C(int a, int b){
	return fac[a] * rfac[b] % Mod * rfac[a - b] % Mod;
}

int main(){
	scanf("%d", &n);
	For(i, 2, n){
		int u, v;
		scanf("%d%d", &u, &v);
		G[u].pb(v), G[v].pb(u);
	}

	fac[0] = 1;
	For(i, 1, n) fac[i] = fac[i - 1] * i % Mod;
	rfac[n] = Pow(fac[n], Mod - 2);
	Forr(i, n - 1, 0) rfac[i] = rfac[i + 1] * (i + 1) % Mod;
	DFS(1, 0);

	For(i, 1, n) coe[i] = coe[i] * fac[i] % Mod;
	For(i, 0, n) F[i] = rfac[n - i];

	NTT(coe, F, 2 * n);

	For(i, 1, n) printf("%lld\n", (1ll * n * C(n, i) % Mod - coe[n + i] * rfac[i] % Mod + Mod) % Mod);
	return 0;
}