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 ll long long
#define cri const register int
#define re register
using namespace std;
const int mod=924844033;
inline int mo(cri x){
	return x>=mod?x-mod:x;
}
inline int qpow(int a,int b,int ans=1){
	for(;b;b>>=1,a=1ll*a*a%mod) if(b&1) ans=1ll*ans*a%mod;
	return ans;
}
vector<int>wn[25];
int v[6000010];
inline void ntt(int *A,cri bin,cri opt){
	for(int i=0;i<bin;i++) if(i<v[i]) swap(A[i],A[v[i]]);
	for(int i=1,t=0;i<bin;i<<=1,t++)
		for(int j=0;j<bin;j+=i<<1)
			for(int k=0;k<i;k++){
				cri x=A[j+k],y=1ll*wn[t][k]*A[j+k+i]%mod;
				A[j+k]=mo(x+y);
				A[j+k+i]=mo(x-y+mod);
			}
	if(opt==-1){
		reverse(A+1,A+bin);
		for(int i=0,pw=qpow(bin,mod-2);i<bin;i++) A[i]=1ll*A[i]*pw%mod;
	}
}
vector<int>to[200010];
int n,A[600010],B[600010],sz[200010],fac[200010],inv[200010];
void dfs(cri x,cri ff){
	sz[x]=1;
	for(int y:to[x]) if(y!=ff){
		dfs(y,x);
		sz[x]+=sz[y];
		A[sz[y]]=mo(A[sz[y]]+mod-1);
	}
	A[n-sz[x]]=mo(A[n-sz[x]]+mod-1);
	A[n-1]=mo(A[n-1]+1);
}
int main(){
	int bin=1;
	scanf("%d",&n);
	while(bin<n+n+1) bin<<=1;
	for(int i=0;i<bin;i++) v[i]=v[i>>1]>>1|(i&1?bin>>1:0);
	for(int i=1,t=0;i<bin;i<<=1,t++){
		wn[t].resize(i);
		wn[t][0]=1;
		cri w=qpow(5,(mod-1)/(i<<1));
		for(int j=1;j<i;j++) wn[t][j]=1ll*wn[t][j-1]*w%mod;
	}
	for(int i=fac[0]=1;i<=n;i++) fac[i]=1ll*fac[i-1]*i%mod;
	inv[n]=qpow(fac[n],mod-2);
	for(int i=n;i;i--) inv[i-1]=1ll*inv[i]*i%mod;
	for(int i=1,x,y;i<n;i++){
		scanf("%d%d",&x,&y);
		to[x].push_back(y);
		to[y].push_back(x);
	}
	dfs(1,0);
	for(int i=1;i<=n;i++) A[i]=1ll*A[i]*fac[i]%mod;
	for(int i=0;i<=n;i++) B[i]=inv[i];
	reverse(B,B+n+1);
	ntt(A,bin,1);ntt(B,bin,1);
	for(int i=0;i<bin;i++) A[i]=1ll*A[i]*B[i]%mod;
	ntt(A,bin,-1);
	for(int i=1;i<=n;i++) printf("%lld\n",(1ll*A[i+n]*inv[i]+1ll*fac[n]*inv[i]%mod*inv[n-i]%mod*i)%mod);
}