// 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: 6.500977242989321x):
#include<bits/stdc++.h>

#define maxn 800010

#define P 924844033

#define G 5

using namespace std;

typedef long long ll;

template<typename T> inline void read(T &x)

{

    x=0;char c=getchar();bool flag=false;

    while(!isdigit(c)){if(c=='-')flag=true;c=getchar();}

    while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}

    if(flag)x=-x;

}

ll n;

ll rev[maxn],f[maxn],g[maxn],fac[maxn],ifac[maxn],siz[maxn],cnt[maxn];

struct edge

{

    int to,nxt;

}e[maxn];

int head[maxn],edge_cnt;

void add(int from,int to)

{

    e[++edge_cnt]={to,head[from]},head[from]=edge_cnt;

}

ll qp(ll x,ll y)

{

    ll v=1;

    while(y)

    {

        if(y&1) v=v*x%P;

        x=x*x%P,y>>=1;

    }

    return v;

}

int calc(int n)

{

    int lim=1;

    while(lim<=n) lim<<=1;

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

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

    return lim;

}

void NTT(ll *a,int lim,int type)

{

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

        if(i<rev[i])

            swap(a[i],a[rev[i]]);

    for(int len=1;len<lim;len<<=1)

    {

        ll wn=qp(G,(P-1)/(len<<1));

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

        {

            ll w=1;

            for(int j=i;j<i+len;++j,w=w*wn%P)

            {

                ll x=a[j],y=w*a[j+len]%P;

                a[j]=(x+y)%P,a[j+len]=(x-y+P)%P;

            }

        }

    }

    if(type==1) return;

    ll inv=qp(lim,P-2);

    for(int i=0;i<lim;++i) a[i]=a[i]*inv%P;

    reverse(a+1,a+lim);

}

void mul(ll *f,ll *g)

{

    int lim=calc(n<<1);

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

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

    NTT(f,lim,-1);

}

void dfs(int x,int fa)

{

    siz[x]=1;

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

    {

        int y=e[i].to;

        if(y==fa) continue;

        dfs(y,x),siz[x]+=siz[y];

    }

    cnt[siz[x]]++,cnt[n-siz[x]]++;

}

ll C(ll n,ll m)

{

    return fac[n]*ifac[m]%P*ifac[n-m]%P;

}

void init()

{

    fac[0]=ifac[0]=1;

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

    ifac[n]=qp(fac[n],P-2);

    for(int i=n-1;i;--i) ifac[i]=ifac[i+1]*(i+1)%P;

}

int main()

{

    read(n),init();

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

    {

        int x,y;

        read(x),read(y);

        add(x,y),add(y,x);

    }

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

    for(int i=1;i<=n;++i) f[i]=cnt[n-i]*fac[n-i]%P,g[i]=ifac[i];

    mul(f,g),reverse(f,f+n+1);

    for(int i=1;i<=n;++i) printf("%lld\n",(n*C(n,i)%P-ifac[i]*f[i]%P+P)%P);

    return 0;

}