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>
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
#define ll long long

using namespace std;

const int N=(1<<20)+100,mod=924844033;
int n,x,y,cnt,head[N],sz[N],fac[N],inv[N],a[N],b[N],w[N],r[N],ans[N],invn;
struct edge{int to,nxt;}e[N<<1];

void adde(int x,int y){
	e[++cnt].to=y; e[cnt].nxt=head[x]; head[x]=cnt;
}

void dfs(int u,int par){
	sz[u]=1;
	for (int i=head[u],v;i;i=e[i].nxt)
		if (v=e[i].to,v!=par) dfs(v,u),sz[u]+=sz[v];
}

int ksm(int x,int y){
	int s=1;
	for (;y;y>>=1,x=(ll)x*x%mod) if (y&1) s=(ll)s*x%mod;
	return s;
}

void init(int n){
	fac[0]=1;
	rep (i,1,n) fac[i]=(ll)fac[i-1]*i%mod;
	inv[n]=ksm(fac[n],mod-2);
	for (int i=n-1;~i;i--) inv[i]=(ll)inv[i+1]*(i+1)%mod;
}

int C(int n,int m){return (ll)fac[n]*inv[m]%mod*inv[n-m]%mod;}

void fft_init(int &n){
	int len=0; while (1<<len<n) len++; n=1<<len;
	int G=ksm(5,(mod-1)>>len); invn=ksm(n,mod-2);
	w[0]=w[n]=1;
	rep (i,1,n-1) w[i]=(ll)w[i-1]*G%mod,r[i]=(r[i>>1]>>1)|((i&1)<<len-1);
}

void fft(int *a,int all,int fl){
	rep (i,0,all-1) if (i<r[i]) swap(a[i],a[r[i]]);
	int n=2,m=1,x=all>>1;
	for (;n<=all;m=n,n<<=1,x>>=1)
		for (int i=0;i<all;i+=n)
			for (int k=0;k<m;k++){
				int t=(ll)w[fl?all-x*k:x*k]*a[i+m+k]%mod;
				a[i+m+k]=(a[i+k]+mod-t)%mod;
				a[i+k]=(a[i+k]+t)%mod;
			}
	if (fl) rep (i,0,all-1) a[i]=(ll)a[i]*invn%mod;
}

void solve(){
	memset(a,0,sizeof(a)),memset(b,0,sizeof(b));
	rep (i,2,n) a[sz[i]]++;
	rep (i,0,n) a[i]=(ll)a[i]*fac[i]%mod,b[i]=inv[i];
	reverse(b,b+n+1);
	int _n=2*n+1; fft_init(_n);
	fft(a,_n,0); fft(b,_n,0);
	rep (i,0,_n-1) a[i]=(ll)a[i]*b[i]%mod;
	fft(a,_n,1);
	rep (i,1,n) ans[i]=(ans[i]+mod-(ll)a[n+i]*inv[i]%mod)%mod;
}

int main(){
	scanf("%d",&n);
	rep (i,1,n-1) scanf("%d%d",&x,&y),adde(x,y),adde(y,x);
	dfs(1,0); init(n);
	rep (i,1,n) ans[i]=(ll)n*C(n,i)%mod;
	solve();
	rep (i,1,n) sz[i]=n-sz[i];
	solve();
	rep (i,1,n) printf("%d\n",ans[i]);
	return 0;
}
/*
这玩意为啥和998244353长那么像啊。。。
*/