// Original C++ solution (slower):
#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;

}

// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>

using namespace std;



inline int read()

{

	int x=0,f=1,c=getchar();

	while(c<48) c=='-'&&(f=-1),c=getchar();

	while(c>47) x=x*10+c-'0',c=getchar();

	return x*f;

}



const int MAXN = 600005;

const int mod = 924844033;

int fac[MAXN],ifac[MAXN],cnt[MAXN],sz[MAXN];

int f[MAXN],g[MAXN],rev[MAXN];

int n;



inline int qpow(int x,int k)

{

	int res=1;

	for(int i=k; i; i>>=1,x=1ll*x*x%mod)

		if(i&1) res=1ll*res*x%mod;

	return res;

}



struct edge {int v,nxt;} e[MAXN<<1];

int head[MAXN];

inline void addEdge(int u,int v)

{

	static int c=0;

	e[++c]=(edge) {v,head[u]},head[u]=c;

}





inline void init(int n)

{

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

	ifac[n]=qpow(fac[n],mod-2);

	for(int i=n; i; --i) ifac[i-1]=1ll*ifac[i]*i%mod;

}



inline int C(int n,int m)

{return 1ll*fac[n]*ifac[m]%mod*ifac[n-m]%mod;}





inline void dfs(int u,int fa)

{

	sz[u]=1;

	for(int i=head[u]; i; i=e[i].nxt)

	{

		int v=e[i].v; if(v==fa) continue;

		dfs(v,u),sz[u]+=sz[v],++cnt[sz[v]];

	}

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

}



inline void ntt(int *a,int n,int op)

{

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

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

	{

		int rot=qpow(op==1?5:554906420,(mod-1)/(i<<1));

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

			for(int k=0,w=1; k<i; ++k,w=1ll*w*rot%mod)

			{

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

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

			}

	}

	if(op==-1)

	{

		int inv=qpow(n,mod-2);

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

	}

}



int main()

{

	n=read();

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

	{

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

		addEdge(u,v),addEdge(v,u);

	}

	dfs(1,0); init(n);

	for(int i=1; i<=n; ++i) f[i]=1ll*cnt[i]*fac[i]%mod;

	reverse(f,f+n+1);

	for(int i=0; i<=n; ++i) g[i]=ifac[i];

	int lim=1,l=0;

	for(; lim<=n<<1; lim<<=1,++l);

	for(int i=0; i<lim; ++i) rev[i]=(rev[i>>1]>>1)|((i&1)<<(l-1));

	ntt(f,lim,1),ntt(g,lim,1);

	for(int i=0; i<lim; ++i) f[i]=1ll*f[i]*g[i]%mod;

	ntt(f,lim,-1);

	reverse(f,f+n+1);

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

		printf("%lld\n",(1ll*n*C(n,i)%mod

		                 -1ll*ifac[i]*f[i]%mod+mod)%mod);

	return 0;

}

// Fastest known solution for this problem: