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 N 800005
#define p 924844033
#define G POW(5)
#define int long long
using namespace std;

inline void rd(int &x){
	x=0;char ch=0;int w=0;
	while(!isdigit(ch)) ch=getchar(),w|=ch=='-';
	while( isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
	x=w?-x:x;
}

int n,lim;
vector<int> g[N];
int A[N],B[N],c[N],r[N];
int f[N],inv[N],ans[N],sz[N];

inline int POW(int a,int b=p-2,int ans=1){
	for(;b;b>>=1,a=a*a%p)
		if(b&1) ans=ans*a%p;
	return ans;
}
void NTT(int *f,int o){
    for(int i=1;i<lim;i++)
        if(i<r[i]) swap(f[i],f[r[i]]);
    for(int len=1;len<lim;len<<=1){
        int w=POW(o,(p-1)/(len<<1));
        for(int b=1,k=0;k<lim;k+=(len<<1),b=1)
            for(int i=k;i<k+len;i++,b=b*w%p){
                int now=f[i+len]*b%p;
                f[i+len]=(f[i]-now+p)%p;f[i]=(f[i]+now)%p;
            }
    }
}
void add(int x,int y){
	g[x].push_back(y);g[y].push_back(x);
}
void mul(int *a,int *b,int n){
	for(lim=1;lim<=n+n+1;lim<<=1);	
    for(int i=1;i<lim;i++)
        r[i]=(r[i>>1]>>1)|((i&1)?lim>>1:0);
    for(int i=0;i<lim;i++)
		A[i]=a[i]*(i<=n),B[i]=b[i]*(i<=n);
    NTT(A,5);NTT(B,5);
    for(int i=0;i<lim;i++)
    	b[i]=A[i]*B[i]%p;
    NTT(b,G);int inv=POW(lim);
    for(int i=0;i<lim;i++)
    	b[i]=b[i]*inv%p;
}
void dfs(int x,int fa=0){
	sz[x]=1;
	for(int y:g[x]) if(y!=fa)
		dfs(y,x),sz[x]+=sz[y];
	c[sz[x]]++;c[n-sz[x]]++;
}
signed main(){
	rd(n);f[0]=inv[0]=ans[0]=1;
	for(int i=1;i<=n;i++)
		f[i]=f[i-1]*i%p,inv[i]=POW(f[i]);
	for(int x,y,i=1;i<n;i++)
		rd(x),rd(y),add(x,y);
	dfs(1);c[n]--;
	for(int i=1;i<=n;i++)
		c[i]=c[i]*f[i]%p,ans[i]=inv[i];
	reverse(c,c+1+n);mul(c,ans,n);reverse(ans,ans+1+n);
	for(int i=1;i<=n;i++)
		ans[i]=ans[i]*inv[i]%p;
	for(int i=1;i<=n;i++)
		printf("%lld\n",(f[n]*inv[i]%p*inv[n-i]%p*n%p+p-ans[i])%p);
}