// Slower C++ solution:
#include <bits/stdc++.h>



using namespace std;



const int P = 924844033;

const int N = 200000 + 5;

const int MAX = 524288 + 5;



int n;



vector <int> G[N];

int siz[N];



int f1[MAX];

int f2[MAX];



int a[N];



int fac[N];

int inv[N];



void dfs(int u, int fa) {

  siz[u] = 1;

  for (int i = 0; i < (int) G[u].size(); ++i) {

    int v = G[u][i];

    if (v == fa) {

      continue;

    }

    dfs(v, u);

    --a[siz[v]];

    siz[u] += siz[v];

  }

  --a[n - siz[u]];

}



int lim;

int bit;



int rev(int x) {

  int ret = 0;

  for (int i = 0; i < bit; ++i) {

    ret <<= 1;

    ret |= x & 1;

    x >>= 1;

  }

  return ret;

}



int revbit[MAX];

int g[MAX];



void fft(int *f, int bit, int o) {

  for (int i = 0; i < lim; ++i) {

    if (revbit[i] > i) {

      swap(f[revbit[i]], f[i]);

    }

  }

  for (int i = 0; i < bit; ++i) {

    int len = 1 << i;

    for (int j = 0; j < lim; j += 1 << i + 1) {

      int wn = 0;

      for (int k = j; k < j + len; ++k) {

        int t = (long long) f[k + len] * g[wn] % P;

        f[k + len] = (f[k] - t) % P;

        if (f[k + len] < 0) {

          f[k + len] += P;

        }

        f[k] = (f[k] + t) % P;

        wn += o * (1 << bit - i - 1);                        

        if (wn < 0) {

          wn += lim;

        }

      }

    }

  }

}



int modexp(int a, int x, int P) {

  int ret = 1;

  while (x) {

    if (x & 1) {

      ret = (long long) ret * a % P;

    }

    a = (long long) a * a % P;

    x >>= 1;

  }

  return ret;

}



void calc() {

  while ((1 << bit) < n + n) {

    ++bit;   

  }

  lim = 1 << bit;

  for (int i = 0; i < lim; ++i) {

    revbit[i] = rev(i);

  }

  g[0] = 1;

  g[1] = modexp(5, (P - 1) / lim, P);

  for (int i = 2; i < lim; ++i) {

    g[i] = (long long) g[i - 1] * g[1] % P;

  }

  fft(f1, bit, 1);

  fft(f2, bit, 1);

  for (int i = 0; i < lim; ++i) {

    f1[i] = (long long) f1[i] * f2[i] % P;

  }

  fft(f1, bit, -1);

  int invlim = modexp(lim, P - 2, P);

  for (int i = 0; i < lim; ++i) {

    f1[i] = (long long) f1[i] * invlim % P;

  }    

}



int main() {

  scanf("%d", &n);

  for (int i = 0, u, v; i < n - 1; ++i) {

    scanf("%d %d", &u, &v);

    --v;

    --u;

    G[u].push_back(v);

    G[v].push_back(u);

  }

  dfs(0, -1);

  a[n] = n;

  fac[0] = 1;

  for (int i = 1; i <= n; ++i) {

    fac[i] = (long long) fac[i - 1] * i % P;

  }

  inv[n] = modexp(fac[n], P - 2, P);

  for (int i = n - 1; i >= 0; --i) {

    inv[i] = (long long) inv[i + 1] * (i + 1) % P;

  }

  for (int i = 1; i <= n; ++i) {

    f1[n - i] = (long long) a[i] * fac[i] % P;

  }

  for (int i = 0; i <= n - 1; ++i) {

    f2[i] = inv[i];

  }

  calc();

  for (int i = n - 1; i >= 0; --i) {

    printf("%lld\n", (long long) f1[i] * inv[n - i] % P);

  }                  

  return 0;

}

// Faster C++ solution for the same problem (measured speedup: 6.500913306175544x):
#include<cstdio>

#include<cstring>

#include<cmath>

#include<algorithm>

#include<vector>

#include<iostream>

#include<map>

#include<string>



#define maxn 400005

#define MOD 924844033



using namespace std;



inline int getint()

{

	int num=0,flag=1;char c;

	while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;

	while(c>='0'&&c<='9')num=num*10+c-48,c=getchar();

	return num*flag;

}



int n;

int fir[maxn],nxt[maxn],to[maxn],cnt;

int rev[maxn<<2];

int sz[maxn],a[maxn<<2],b[maxn<<2];

int fac[maxn],inv[maxn];



inline void newnode(int u,int v)

{to[++cnt]=v,nxt[cnt]=fir[u],fir[u]=cnt;}

inline void dfs(int u,int fa)

{

	sz[u]=1;

	for(int i=fir[u];i;i=nxt[i])if(to[i]!=fa)

	{

		dfs(to[i],u),sz[u]+=sz[to[i]];

		a[sz[to[i]]]++;

	}

	a[n-sz[u]]++;

}



inline int ksm(int num,int k)

{

	int ret=1;

	for(;k;k>>=1,num=1ll*num*num%MOD)if(k&1)ret=1ll*ret*num%MOD;

	return ret;

}

inline void getrev(int N)

{for(int i=0;i<N;i++)rev[i]=(rev[i>>1]>>1)|(i&1?N>>1:0);}

inline void NTT(int *a,int opt,int N)

{

	for(int i=0;i<N;i++)if(rev[i]<i)swap(a[i],a[rev[i]]);

	for(int i=1;i<N;i<<=1)

	{

		int wn=ksm(5,(MOD-1)/(i<<1));

		if(!~opt)wn=ksm(wn,MOD-2);

		for(int j=0;j<N;j+=(i<<1))for(int k=0,w=1;k<i;k++,w=1ll*w*wn%MOD)

		{

			int x=a[j+k],y=1ll*a[i+j+k]*w%MOD;

			a[j+k]=(x+y)%MOD,a[i+j+k]=(x-y+MOD)%MOD;

		}

	}

	if(!~opt)for(int i=0,Inv=ksm(N,MOD-2);i<N;i++)a[i]=1ll*a[i]*Inv%MOD;

}



int main()

{

	n=getint();

	for(int i=1;i<n;i++)

	{

		int u=getint(),v=getint();

		newnode(u,v),newnode(v,u);

	}

	fac[0]=fac[1]=inv[0]=inv[1]=1;

	for(int i=2;i<=n;i++)fac[i]=1ll*fac[i-1]*i%MOD;

	for(int i=2;i<=n;i++)inv[i]=1ll*inv[MOD%i]*(MOD-MOD/i)%MOD;

	for(int i=2;i<=n;i++)inv[i]=1ll*inv[i]*inv[i-1]%MOD;

	dfs(1,0);a[0]=0;

	for(int i=0;i<=n;i++)a[i]=1ll*a[i]*fac[i]%MOD;

	for(int i=0;i<=n;i++)b[i]=inv[n-i];

	int len=1;

	while(len<=2*n)len<<=1;

	getrev(len);

	NTT(a,1,len),NTT(b,1,len);

	for(int i=0;i<len;i++)a[i]=1ll*a[i]*b[i]%MOD;

	NTT(a,-1,len);

	for(int i=1;i<=n;i++)

	{

		int ans=1ll*fac[n]*inv[i]%MOD*inv[n-i]%MOD*n%MOD;

		printf("%lld\n",(ans-1ll*a[i+n]*inv[i]%MOD+MOD)%MOD);

	}

}