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;
#define FIO "A"
const int N=6e5,MOD=924844033,G=5;

int n,l=1,ans[N],u,v,sz[N],a[N],b[N],c[N],d[N],rev[N],w[2][N];
int fac[N],inv[N],invc[N];
int h[N],to[N],nxt[N],ecnt;

inline void adde(int u,int v){nxt[++ecnt]=h[u];h[u]=ecnt;to[ecnt]=v;}

inline int add(int a,const int &b){if((a+=b)>=MOD)a-=MOD;return a;}
inline int sub(int a,const int &b){if((a-=b)<0)a+=MOD;return a;}
inline int mul(const int &a,const int &b){return 1ll*a*b%MOD;}
inline int qpow(int a,int b){int ret=1;for(;b;b>>=1,a=mul(a,a))if(b&1)ret=mul(ret,a);return ret;}
inline int C(const int &a,const int &b){return mul(mul(fac[a],invc[b]),invc[a-b]);}

inline void dfs(int u,int fa){
	sz[u]=1;
	for(int i=h[u],v;i;i=nxt[i])if((v=to[i])^fa){
		dfs(v,u);
		c[sz[v]]++;
		sz[u]+=sz[v];
	}
	c[n-sz[u]]++;
}

inline void fft(int *a,int opt){
	for(int i=0;i<l;i++)if(i<rev[i])swap(a[i],a[rev[i]]);
	for(int i=1;i<l;i<<=1)
		for(int j=0;j<l;j+=i<<1)
			for(int k=0;k<i;k++){
				int x=a[j+k],y=mul(a[i+j+k],w[opt][i+k]);
				a[j+k]=add(x,y);
				a[i+j+k]=sub(x,y);
			}
	if(opt)for(int i=0,inv=qpow(l,MOD-2);i<l;i++)a[i]=mul(a[i],inv);
}

int main(){
	scanf("%d",&n);
	while(l<=n<<1)l<<=1;
	for(int i=0;i<l;i++)rev[i]=(rev[i>>1]>>1)|((i&1)*(l>>1));
	for(int i=1;i<l;i<<=1){
		w[0][i]=w[1][i]=1;
		int wn1=qpow(G,(MOD-1)/(i<<1)),wn0=qpow(wn1,MOD-2);
		for(int j=1;j<i;j++)w[0][i+j]=mul(w[0][i+j-1],wn0),w[1][i+j]=mul(w[1][i+j-1],wn1);
	}
	for(int i=1;i<n;i++)scanf("%d%d",&u,&v),adde(u,v),adde(v,u);
	inv[0]=inv[1]=fac[0]=fac[1]=invc[0]=invc[1]=1;
	for(int i=2;i<=n;i++)fac[i]=mul(fac[i-1],i),inv[i]=mul(MOD-inv[MOD%i],MOD/i),invc[i]=mul(invc[i-1],inv[i]);
	dfs(1,0);
	for(int i=0;i<=n;i++)a[i]=mul(c[i],fac[i]),b[n-i]=invc[i];
	/*
	for(int i=1;i<=n;i++)printf("%d%c",a[i],i==n?'\n':' ');
	for(int i=1;i<=n;i++)printf("%d%c",b[i],i==n?'\n':' ');
	for(int i=1;i<=n;i++)printf("%d%c",c[i],i==n?'\n':' ');
	*/
	fft(a,0);fft(b,0);
	for(int i=0;i<l;i++)ans[i]=mul(a[i],b[i]);
	fft(ans,1);
	for(int i=1;i<=n;i++)printf("%d\n",sub(mul(C(n,i),n),mul(invc[i],ans[n+i])));
	return 0;
}
