// 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: 6.476715045866356x):
#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <string>

#include <algorithm>

#define N 601000

typedef long long ll; 

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

  x = 0; char  c = getchar();

  while (!isdigit(c)) c = getchar();

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

}

using namespace std;

const int P = 924844033;

const int G = 5;

int Gi; 

inline ll quickpow(ll x, int k) {

  ll res = 1;

  while (k) {

    if (k & 1)  res = res * x % P;  

    x = x *x  % P;

    k >>= 1;

  }

  return res;

}

int n;

struct edge{

  int nxt, to; 

}e[N << 1]; 

int head[N], ecnt;

inline void addedge(int from, int to) {

  e[++ecnt] = (edge){head[from], to};

  head[from] = ecnt;

}



int siz[N], dep[N], fa[N];

void dfs_son(int cur, int faa) {

  siz[cur] = 1, fa[cur] = faa;

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

    int to = e[i].to; if (to == faa)  continue;

    dfs_son(to, cur);

    siz[cur] += siz[to];

  }

}



int cnt[N];

ll f[N], g[N], A[N], ans[N];

int limi, L, r[N];

inline void ntt(ll *a, int type) {

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

    if (i < r[i]) swap(a[i], a[r[i]]);

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

    ll T = quickpow(type == 1 ? G : Gi, (P - 1) / (i << 1));

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

      ll t = 1;

      for (int k = 0; k < i; ++k, t = t * T % P) {

        ll nx = a[j + k], ny = t * a[j + k + i] % P;

        a[j + k] = (nx + ny) % P;

        a[j + k + i] = (nx - ny + P) % P;

      }   

    }   

  }

  if (type == -1) {

    ll inv = quickpow(limi, P - 2); 

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

  }

}



ll jie[N], jieni[N];

inline ll get_c(int n, int m) {

  if (n < 0 || m < 0 || n - m < 0)  return 0;

  return jie[n] * jieni[m] % P * jieni[n - m] % P;

}

inline void work() {

  limi = 1;

  while (limi <= n + n + 2) limi <<= 1, ++L;

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

    r[i] = (r[i >> 1] >> 1) | ((i & 1) << (L - 1));

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

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

  ntt(A, -1);

//  for (int i= 0; i <= n; ++i)	printf("A(yuan)[%d] = %lld\n", i, A[n - i]);

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

    ll ans = (1ll * n * get_c(n, k) % P - jieni[k] * A[n - k] % P) % P;

//    printf("%d * %lld - %lld * %lld = \n", n, get_c(n, k), jie[k], A[n - k]);

    printf("%lld\n", (ans % P + P) % P); 

  }

}



inline void init() {

  jie[0] = jieni[0] = 1;

  int up = n + 10; 

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

  jieni[up] = quickpow(jie[up], P - 2); 

  for (int i = up - 1; i; --i)  jieni[i] = jieni[i + 1] * (i + 1) % P;

  Gi = quickpow(G, P - 2); 

}



int main() {

  read(n); init();

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

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

  }

  dfs_son(1, 0); 

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

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

  }

  for (int i = 0; i <= n; ++i)  f[i] = 1ll * cnt[n - i] * jie[n - i] % P;

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

//  for (int i = 0; i <= n; ++i)	printf("f[%d] = %lld, g = %lld\n", i, f[i], g[i]);

  work();

  return 0;

}



/*

3

1 2

2 3

//3 7 3

*/