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 <iostream>
#include <algorithm>

using namespace std;

const int N=1000010,P=924844033;

int n,cnt,G[N],a[N],A[N],B[N];
struct edge{
  int t,nx;
}E[N<<1];

inline void addedge(int x,int y){
  E[++cnt].t=y; E[cnt].nx=G[x]; G[x]=cnt;
  E[++cnt].t=x; E[cnt].nx=G[y]; G[y]=cnt;
}

int dfs(int x,int f){
  int ret=1;
  for(int i=G[x];i;i=E[i].nx){
    if(E[i].t==f) continue;
    int cur=dfs(E[i].t,x);
    a[cur]++; ret+=cur;
  }
  a[n-ret]++;
  return ret;
}

int w[2][N],rev[N],L,m,fac[N],inv[N];

inline int Pow(int x,int y){
  int ret=1;
  for(;y;y>>=1,x=1LL*x*x%P) if(y&1) ret=1LL*ret*x%P;
  return ret;
}

inline void Pre(){
  int g=Pow(5,(P-1)/m);
  w[0][0]=w[1][0]=1;
  for(int i=1;i<m;i++) w[1][i]=1LL*w[1][i-1]*g%P;
  for(int i=1;i<m;i++) w[0][i]=w[1][m-i];

  fac[0]=1; for(int i=1;i<m;i++) fac[i]=1LL*fac[i-1]*i%P;
  inv[1]=1; for(int i=2;i<m;i++) inv[i]=1LL*(P-P/i)*inv[P%i]%P;
  inv[0]=1; for(int i=1;i<m;i++) inv[i]=1LL*inv[i]*inv[i-1]%P;
}

inline void NTT(int *a,int m,int r){
  for(int i=1;i<m;i++) if(rev[i]>i) swap(a[i],a[rev[i]]);
  for(int i=1;i<m;i<<=1)
    for(int j=0;j<m;j+=i<<1)
      for(int k=0;k<i;k++){
	int x=a[j+k],y=1LL*w[r][m/(i<<1)*k]*a[j+k+i]%P;
	a[j+k]=(x+y)%P; a[j+k+i]=(x+P-y)%P;
      }
  if(!r) for(int i=0,inv=Pow(m,P-2);i<m;i++) a[i]=1LL*a[i]*inv%P;
}

inline int C(int x,int y){
  return 1LL*fac[x]*inv[y]%P*inv[x-y]%P;
}

int main(){
  scanf("%d",&n);
  for(int i=1,x,y;i<n;i++)
    scanf("%d%d",&x,&y),addedge(x,y);
  dfs(1,0);
  for(m=1;m<=n;m<<=1,L++); m<<=1; Pre();
  for(int i=1;i<m;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<L);
  for(int i=1;i<=n;i++) A[i]=1LL*a[i]*fac[i]%P,B[m-i]=inv[i]; B[0]=1;
  NTT(A,m,1); NTT(B,m,1);
  for(int i=0;i<m;i++) A[i]=1LL*A[i]*B[i]%P;
  NTT(A,m,0);
  for(int i=1;i<=n;i++){
    int cur=1LL*n*C(n,i)%P;
    cur=(cur-1LL*inv[i]*A[i])%P;
    printf("%d\n",(cur+P)%P);
  }
  return 0;
}
