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 <algorithm>
#include <cstdlib>
#include <queue>
using namespace std;
#define mod 924844033
typedef long long ll;
#define N 400050
int head[N],to[N],nxt[N],cnt,siz[N],pp[N],n;
ll fac[N],inv[N];
inline void add(int u,int v) {
	to[++cnt]=v; nxt[cnt]=head[u]; head[u]=cnt;
}
void dfs(int x,int y) {
	int i; siz[x]=1; 
	for(i=head[x];i;i=nxt[i]) if(to[i]!=y) {
		dfs(to[i],x);
		siz[x]+=siz[to[i]];
		pp[siz[to[i]]]++;
	}
	if(x!=1) pp[n-siz[x]]++;
}
ll qp(ll x,ll y) {
	ll re=1; for(;y;y>>=1,x=x*x%mod) if(y&1) re=re*x%mod; return re;
}
ll INV(ll x) {return qp(x,mod-2);}
void ntt(ll *a,int len,int flg) {
	int i,j,k,t; ll tmp,w,wn;
	for(i=k=0;i<len;i++) {
		if(i>k) swap(a[i],a[k]);
		for(j=len>>1;(k^=j)<j;j>>=1) ;
	}
	for(k=2;k<=len;k<<=1) {
		t=k>>1; wn=qp(5,(mod-1)/k); if(flg==-1) wn=INV(wn);
		for(i=0;i<len;i+=k) {
			w=1;
			for(j=i;j<i+t;j++) {
				tmp=a[j+t]*w%mod;
				a[j+t]=(a[j]-tmp)%mod;
				a[j]=(a[j]+tmp)%mod;
				w=w*wn%mod;
			}
		}
	}
	if(flg==-1) for(tmp=INV(len),i=0;i<len;i++) a[i]=a[i]*tmp%mod;
}
ll a[N<<1],b[N<<1],g[N<<1],f[N<<1];
int main() {
	scanf("%d",&n);
	int i,x,y;
	for(i=1;i<n;i++) scanf("%d%d",&x,&y),add(x,y),add(y,x);
	dfs(1,0);
	for(fac[0]=i=1;i<=n;i++) fac[i]=fac[i-1]*i%mod;
	inv[n]=INV(fac[n]);
	for(i=n-1;i>=0;i--) inv[i]=inv[i+1]*(i+1)%mod;
	for(i=0;i<=n+1;i++) a[i]=pp[i]*fac[i]%mod, b[i]=inv[n-i+1];
	int len=1;
	for(;len<=(n+n);len<<=1) ;
	ntt(a,len,1); ntt(b,len,1);
	for(i=0;i<len;i++) a[i]=a[i]*b[i]%mod;
	ntt(a,len,-1);
	for(i=1;i<=n;i++) g[i]=a[n+i+1];
	for(i=1;i<=n;i++) {
		f[i]=(n*fac[n]%mod*inv[n-i]%mod-g[i])*inv[i]%mod;
		printf("%lld\n",(f[i]+mod)%mod);
	}
}