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 <cstring>
#include <cstdlib>
#include <algorithm>
#include <vector>
#define LL long long
using namespace std;
const LL Maxn = 200010;
const LL Mod = 924844033;
const LL G = 5;
LL A[Maxn<<2], B[Maxn<<2], N;
LL size[Maxn], n;
LL jc[Maxn<<2], inv[Maxn<<2], invN;
vector <LL> vec[Maxn];
void dfs(LL x, LL fa) {
	LL sz = vec[x].size();
	size[x] = 1;
	for(LL i = 0; i < sz; i++){
		LL y = vec[x][i];
		if(y == fa) continue;
		dfs(y, x);
		A[size[y]]++; A[n-size[y]]++;
		size[x] += size[y];
	}
}
LL qpow(LL x, LL k) {
	LL ret = 1;
	while(k){
		if(k&1) ret = ret*x%Mod;
		x = x*x%Mod;
		k >>= 1;
	}
	return ret;
}
void dft(LL *a, LL op) {
	LL i, j, k;
	j = 0;
	for(i = 0; i < N; i++){
		if(i < j) swap(a[i], a[j]);
		k = N>>1;
		while(j&k) j -= k, k >>= 1;
		j += k;
	}
	for(i = 2; i <= N; i <<= 1){
		LL wn = qpow(G, op == 1 ? (Mod-1)/i : Mod-1-(Mod-1)/i);
		for(j = 0; j < N; j += i){
			LL w = 1;
			for(k = j; k < j+i/2; k++){
				LL x = a[k], y = a[k+i/2]*w%Mod;
				a[k] = (x+y)%Mod; a[k+i/2] = (x-y+Mod)%Mod;
				w = w*wn%Mod;
			}
		}
	}
	if(op == -1) for(i = 0; i < N; i++) a[i] = a[i]*invN%Mod;
}
LL C(LL x, LL y) { return jc[x]*inv[y]%Mod*inv[x-y]%Mod; }
int main() {
	LL i, j, k;
	scanf("%lld", &n);
	for(i = 1; i < n; i++){
		LL x, y;
		scanf("%lld%lld", &x, &y);
		vec[x].push_back(y); vec[y].push_back(x);
	}
	dfs(1, 0);
	N = 1;
	while(N < 2*n-1) N <<= 1;
	jc[0] = 1;
	for(i = 1; i <= N; i++) jc[i] = jc[i-1]*i%Mod;
	inv[0] = inv[1] =1;
	for(i = 2; i <= N; i++) inv[i] = inv[Mod%i]*(Mod-Mod/i)%Mod;
	invN = inv[N];
	for(i = 2; i <= N; i++) inv[i] = inv[i-1]*inv[i]%Mod;
	for(i = 1; i <= n-1; i++) A[i] = A[i]*jc[i]%Mod;
	for(i = n-1; i >= 0; i--) B[i] = inv[n-1-i];
	dft(A, 1); dft(B, 1);
	for(i = 0; i < N; i++) A[i] = A[i]*B[i]%Mod;
	dft(A, -1);
	for(i = 1; i <= n; i++){
		LL ans = n*C(n, i)%Mod;
		LL minus = inv[i]*A[n+i-1]%Mod;
		printf("%lld\n", (ans-minus+Mod)%Mod);
	}
	return 0;
}