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<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N=2e5+10,M=5e5+10,T=1e6+5,mod=924844033,sqr=5,sqrinv=554906420;
int n,siz[N],limit,lg,E,ot[M],ne[M],g[N];
int fac[N],inv[N],infac[N];
int a[T],b[T],r[T];
inline int ksm(int x,int y){
    int res=1;
    while(y){if(y&1)res=1ll*res*x%mod;x=1ll*x*x%mod; y>>=1;}
    return res;
}
inline void add(int &u,int &v){ot[++E]=v,ne[E]=g[u],g[u]=E;}
inline void init(){
    cin>>n;
    for(int i=1,u,v;i<n;i++){
        scanf("%d%d",&u,&v);
        add(u,v),add(v,u);
    }
}
void dfs(int x,int fa){
    siz[x]=1;
    for(int i=g[x];i;i=ne[i]){
        int to=ot[i];
        if(to==fa)continue;
        dfs(to,x); siz[x]+=siz[to];
        a[siz[to]]++;
    }
    a[n-siz[x]]++;
}
inline int solve_complex(int x,int tp){
    if(~tp)return ksm(sqr,x);
    else return ksm(sqrinv,x);
}
inline void DFT(int *A,int tp){
    for(int i=0;i<limit;i++)if(i<r[i])swap(A[i],A[r[i]]);
    for(int len=1;len<limit;len<<=1){
        int wn=solve_complex((mod-1)/(2*len),tp);
        for(int j=0;j<limit;j+=len<<1){
            int w=1;
            for(int k=0;k<len;++k,w=1ll*w*wn%mod){
                int x=A[j+k],y=1ll*w*A[j+k+len]%mod;
                A[j+k]=x+y,A[j+k+len]=x-y;
                if(A[j+k]>=mod)A[j+k]-=mod;
                if(A[j+k+len]<0)A[j+k+len]+=mod;
            }
        }
    }
}
inline void pre(){
    register int i; dfs(1,0);
    fac[0]=1;for(i=1;i<=n;++i) fac[i]=1ll*fac[i-1]*i%mod;
    inv[1]=1;for(i=2;i<=n;++i) inv[i]=1ll*inv[mod%i]*(mod-mod/i)%mod;
    infac[0]=1;for(i=1;i<=n;++i) infac[i]=1ll*infac[i-1]*inv[i]%mod;
}
inline void work(){
    a[0]=0; register int i;
    for(i=0;i<=n;i++)a[i]=1ll*a[i]*fac[i]%mod;
    for(i=0;i<=n;i++)b[i]=infac[n-i];
    limit=1; while(limit<=2*n)limit<<=1,++lg;
    for(i=1;i<limit;i++)r[i]=((r[i>>1]>>1)|((i&1)<<(lg-1)));
    DFT(a,1),DFT(b,1);
    for(i=0;i<limit;i++)a[i]=1ll*a[i]*b[i]%mod;
    DFT(a,-1);
    int INV=ksm(limit,mod-2);
    for(i=1;i<=n;i++)a[i+n]=1ll*a[i+n]*INV%mod*infac[i]%mod;
    for(i=1;i<=n;i++)a[i]=1ll*fac[n]*infac[i]%mod*infac[n-i]%mod*n%mod;
    for(i=1;i<=n;i++){
        a[i]-=a[i+n];
        if(a[i]<0)a[i]+=mod;
    }
    for(i=1;i<=n;i++)printf("%d\n",a[i]);
}
signed main(){
    init();pre();work();return 0;
}