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>
using namespace std;

#define ll long long
#define N (1<<19)
#define mod 924844033
#define gen 5
ll K(ll x,int y=mod-2){
  ll t=1;
  for (;y;y>>=1,x=x*x%mod) if (y&1) t=t*x%mod;
  return t;
}
namespace FFT{
  void fft(int *a,int n){
    for (int i=0,j;i<1<<n;++i){
      for (int o=j=0;o<n;++o) j=j<<1|i>>o&1;
      if (i<j) swap(a[i],a[j]);
    }
    for (int o=0,len,w0;o<n;++o){
      len=1<<o; w0=K(gen,(mod-1)/(len*2));
      for (int i=0;i<1<<n;i+=len*2){
        int *l=a+i,*r=a+i+len,w=1,x,y;
        for (int j=0;j<len;++j){
          x=l[j], y=1ll*r[j]*w%mod;
          l[j]= x+y>=mod? x+y-mod: x+y;
          r[j]= x-y<0? x-y+mod: x-y;
          w=(ll)w*w0%mod;
        }
      }
    }
  }
  void realmain(int *a,int *b,int n){
    fft(a,n); fft(b,n);
    for (int i=0;i<1<<n;++i) a[i]=1ll*a[i]*b[i]%mod;
    reverse(a+1,a+(1<<n));
    fft(a,n); ll inv=K(1<<n);
    for (int i=0;i<1<<n;++i) a[i]=a[i]*inv%mod;
  }
}

int n,a[N],b[N],x,y; ll jc[N],inv[N];
vector<int>G[N/2];

int dfs(int x,int fff){
  int sz=1;
  for (auto y:G[x]) if (y-fff) sz+=dfs(y,x);
  if (fff) ++a[sz],++a[n-sz];
  return sz;
}

int main(){
  cin.tie(0); cin>>n;
  for (int i=n;--i;) cin>>x>>y, G[x].push_back(y), G[y].push_back(x);
  dfs(1,0);
  jc[0]=inv[0]=b[0]=1;
  for (int i=1;i<=n;++i) b[i]=inv[i]=K(jc[i]=jc[i-1]*i%mod);
  for (int i=0;i<=n;++i) a[i]=a[i]*jc[i]%mod;
  reverse(a,a+n+1);
  FFT::realmain(a,b,19);
  reverse(a,a+n+1);
  for (int i=1;i<=n;++i)
    printf("%lld\n",(jc[n]*inv[i]%mod*inv[n-i]%mod*n%mod\
                     -a[i]*inv[i]%mod+mod)%mod);
}