Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=200000+10,maxlen=maxn*4,mo=924844033,GG=5;
int size[maxn],h[maxn],go[maxn*2],nxt[maxn*2];
int w[maxlen],a[maxlen],b[maxlen],tt[maxlen],c[maxlen],rev[maxlen],fac[maxn],inv[maxn];
int i,j,k,l,t,n,m,tot,len,ans,ni;
double ce;

void add(int x,int y)
{
    go[++tot]=y;
    nxt[tot]=h[x];
    h[x]=tot;
}
int qsm(int x,int y){
    if (!y) return 1;
    int t=qsm(x,y/2);
    t=(ll)t*t%mo;
    if (y%2) t=(ll)t*x%mo;
    return t;
}
void dfs(int x,int y){
    a[n]++;
    int t=h[x];
    size[x]=1;
    while (t){
        if (go[t]!=y){
            dfs(go[t],x);
            a[size[go[t]]]--;
            size[x]+=size[go[t]];
        }
        t=nxt[t];
    }
    a[n-size[x]]--;
}

void prepare()
{
  w[0]=1; w[1]=qsm(GG,(mo-1)/len);
  for(int i=2; i<=len; i++) w[i]=(ll)w[i-1]*w[1]%mo;
  for(int i=0; i<len; i++)
	{
    int p=0;
    for(int j=0,tp=i;j<ce;j++,tp/=2) p=(p<<1)+(tp%2);
    rev[i]=p;
  }
  ni=qsm(len,mo-2);
}

void DFT(int *a,int sig)
{
  for(int i=0; i<len; i++) tt[rev[i]]=a[i];
  for(int m=2;m<=len;m*=2)
	{
    int half=m/2,bei=len/m;
    for(int i=0; i<half; i++)
		{
      int wi=sig>0?w[i*bei]:w[len-i*bei];
      for(int j=i;j<len;j+=m)
			{
        int u=tt[j],v=(ll)tt[j+half]*wi%mo;
        tt[j]=(u+v)%mo;
        tt[j+half]=(u-v)%mo;
      }
    }
  }
  if (sig==-1)
    for(int i=0; i<len; i++) tt[i]=(ll)tt[i]*ni%mo;
  for(int i=0; i<len; i++) a[i]=tt[i];
}

void NTT()
{
  DFT(a,1); DFT(b,1);
  for(int i=0; i<len; i++) a[i]=(ll)a[i]*b[i]%mo;
  DFT(a,-1);
  for(int i=0; i<len; i++) c[i]=a[i];
}

int main()
{
  scanf("%d",&n);
  for(int i=1; i<n; i++)
	{
    scanf("%d%d",&j,&k); add(j,k); add(k,j);
  }
  dfs(1,0); fac[0]=1;
  for(int i=1; i<=n; i++) fac[i]=(ll)fac[i-1]*i%mo;
  inv[n]=qsm(fac[n],mo-2);
  for(int i=n-1; i>=0; i--) inv[i]=(ll)inv[i+1]*(i+1)%mo;
  for(int i=0; i<=n; i++) a[i]=(ll)a[i]*fac[i]%mo;
  for(int i=0; i<=n; i++) b[i]=inv[i];
  reverse(b,b+n+1);
  len=1;
  while (len<=2*n) len*=2;
  ce=log(len)/log(2);
  prepare(); NTT();
  for(int i=1; i<=n; i++)
	{
    ans=(ll)inv[i]*c[n+i]%mo;
    (ans+=mo)%=mo;
    printf("%d\n",ans);
  }
  return 0;
}