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<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<vector>
#define ll long long
#define pb push_back
using namespace std;
const int N=200010;
const ll P=924844033;
int n;
vector<int>g[N];
int siz[N];
int rev[N<<2],lim,lev;
ll pwr(int ll x,ll y){
	if(!y)return 1;
	if(y&1)return pwr(x,y-1)*x%P;
	ll z=pwr(x,y/2);
	return z*z%P;
}
inline void ntt(ll *a,int t){
	for(int i=0;i<lim;i++)
		if(i<rev[i])
			swap(a[i],a[rev[i]]);
	for(int mid=1;mid<lim;mid<<=1){
		int rgt=mid<<1;
		ll wn=pwr(5,(P-1)/rgt);
		if(t==-1)wn=pwr(wn,P-2);
		for(int loc=0;loc<lim;loc+=rgt){
			ll w=1;
			for(int k=0;k<mid;k++){
				ll x=a[loc+k],y=w*a[loc+mid+k]%P;
				a[loc+k]=(x+y)%P;
				a[loc+mid+k]=(x-y+P)%P;
				w=w*wn%P;
			}
		}
	}
	if(t==-1){
		ll invlim=pwr(lim,P-2);
		for(int i=0;i<lim;i++)
			a[i]=a[i]*invlim%P;
	}
}
ll a[N<<2],b[N<<2];
ll fac[N],inv[N];
void dfs(int x,int f){
	siz[x]=1;
	int i,k;
	for(i=g[x].size()-1;~i;i--){
		k=g[x][i];
		if(k==f)continue;
		dfs(k,x);
		a[siz[k]]++;
		siz[x]+=siz[k];
	}
	a[n-siz[x]]++;
}
inline void init(){
	fac[0]=1;
	int i;
	for(i=1;i<=n;i++)
		fac[i]=fac[i-1]*i%P;
	inv[n]=pwr(fac[n],P-2);
	for(i=n;i;i--)
		inv[i-1]=inv[i]*i%P;
}
int main(){
	int i;
	scanf("%d",&n);
	for(i=1;i<n;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		g[x].pb(y);
		g[y].pb(x);
	}
	dfs(1,0);
	for(i=1;i<n;i++)
		a[i]=P-a[i];
	a[n]=n;
	init();
	for(i=1;i<=n;i++)
		a[i]=a[i]*fac[i]%P;
	for(i=1;i<=n;i++)
		b[i]=inv[n-i];
	//(i)+(n-(i-k))=(n+k)
	lim=1;
	while(lim<=n*2){
		lim<<=1;
		lev++;
	}
	for(i=0;i<lim;i++)
		rev[i]=(rev[i>>1]>>1)|((i&1)<<(lev-1));
	ntt(a,1);
	ntt(b,1);
	for(i=0;i<lim;i++)
		a[i]=a[i]*b[i]%P;
	ntt(a,-1);
	for(i=1;i<=n;i++)
		printf("%lld\n",a[i+n]*inv[i]%P);
	return 0;
}