// 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: 1.6915196897808362x):
#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

typedef long long lint;

const int N=262144;

const int mod=924844033;

//5

int n,cal[N],fac[N],inv[N];



inline int nxi(){

	int x=0;

	char c;

	while((c=getchar())>'9'||c<'0');

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

	return x;

}



inline int fpow(int x,int t){

	int ans=1;

	for(; t; x=(lint)x*x%mod,t>>=1){

		if(t&1) ans=(lint)ans*x%mod; 

	}

	return ans;

}



inline void init(){

	fac[0]=1;

	for(int i=1; i<N; ++i) fac[i]=(lint)fac[i-1]*i%mod;

	inv[N-1]=fpow(fac[N-1],mod-2);

	for(int i=N-1; i; --i) inv[i-1]=(lint)inv[i]*i%mod;

}



inline int C(const int x,const int y){

	if(x<y) return 0;

	return (lint)fac[x]*inv[y]%mod*inv[x-y]%mod;

}



namespace G{

	int cnt,fir[N],sz[N];

	struct edge{

		int to,nx;

	}eg[N<<1];



	inline void add(const int a,const int b){

		eg[++cnt]=(edge){b,fir[a]};

		fir[a]=cnt;

	}



	inline void dfs_cal(const int x){

		sz[x]=1;

		for(int i=fir[x]; i; i=eg[i].nx){

			const int y=eg[i].to;

			if(!sz[y]){

				dfs_cal(y);

				++cal[sz[y]];

				sz[x]+=sz[y];

			}

		}

		++cal[n-sz[x]];

	}

}



namespace P{



	void ntt(int *a,int len,bool f){

		static int rev[N<<1];

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

			rev[i]=rev[i>>1]>>1|(i&1?len>>1:0);

			if(i<rev[i]) std::swap(a[i],a[rev[i]]);

		}

		for(int i=1; i<len; i<<=1){

			int t=fpow(5,(mod-1)/(i<<1));

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

				for(int w=1,k=0; k<i; ++k,w=(lint)w*t%mod){

					lint p=(lint)w*a[j+k+i];

					a[j+k+i]=(a[j+k]-p)%mod;

					a[j+k]=(a[j+k]+p)%mod;

				}

			}

		}

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

			if(a[i]<0) a[i]+=mod;

		}

		if(f==0) return;

		std::reverse(a+1,a+len);

		int tmp=fpow(len,mod-2);

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

			a[i]=(lint)a[i]*tmp%mod;

		}

	}



	class poly{

		public:

			int len,*n;

			poly(int len=0){

				this->len=len;

				n=new int[len+1];

			}

			friend poly operator * (poly a,poly b);

	}ans;



	poly operator * (poly a,poly b){

		static int buka[N<<1],bukb[N<<1];

		poly c(a.len+b.len);

		int len=1;

		while(len<=c.len) len<<=1;

		memset(buka,0,len*sizeof(int));

		memset(bukb,0,len*sizeof(int));

		memcpy(buka,a.n,(a.len+1)*sizeof(int));

		memcpy(bukb,b.n,(b.len+1)*sizeof(int));

		ntt(buka,len,0);

		ntt(bukb,len,0);

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

			buka[i]=(lint)buka[i]*bukb[i]%mod;

		}

		ntt(buka,len,1);

		memcpy(c.n,buka,(c.len+1)*sizeof(int));

		return c;

	}



	inline void get_poly(){

		poly a(n),b(n);

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

			a.n[i]=(lint)cal[i]*fac[i]%mod;

			b.n[i]=inv[n-i-1];

		}

		ans=a*b;

	}

}



int main(){

	init();

	n=nxi();

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

		const int a=nxi(),b=nxi();

		G::add(a,b);

		G::add(b,a);

	}

	G::dfs_cal(1);

	P::get_poly();

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

		const int all=(lint)n*C(n,i)%mod;

		const int invld=(lint)inv[i]*P::ans.n[n-1+i]%mod;

		printf("%d\n",(all+mod-invld)%mod);

	}

	return 0;

}