One day, Takahashi was given the following problem from Aoki:

* You are given a tree with N vertices and an integer K. The vertices are numbered 1 through N. The edges are represented by pairs of integers (a_i, b_i).
* For a set S of vertices in the tree, let f(S) be the minimum number of the vertices in a subtree of the given tree that contains all vertices in S.
* There are <image> ways to choose K vertices from the trees. For each of them, let S be the set of the chosen vertices, and find the sum of f(S) over all <image> ways.
* Since the answer may be extremely large, print it modulo 924844033(prime).



Since it was too easy for him, he decided to solve this problem for all K = 1,2,...,N.

Constraints

* 2 ≦ N ≦ 200,000
* 1 ≦ a_i, b_i ≦ N
* The given graph is a tree.

Input

The input is given from Standard Input in the following format:


N
a_1 b_1
a_2 b_2
:
a_{N-1} b_{N-1}


Output

Print N lines. The i-th line should contain the answer to the problem where K=i, modulo 924844033.

Examples

Input

3
1 2
2 3


Output

3
7
3


Input

4
1 2
1 3
1 4


Output

4
15
13
4


Input

7
1 2
2 3
2 4
4 5
4 6
6 7


Output

7
67
150
179
122
45
7
#include <bits/stdc++.h>
using namespace std;
const int MOD = 924844033, G = 5, N = 600000;
int rev[N];
int power(int a,int b) {
  int ret = 1;
  while (b) {
    if (b&1) ret = 1ll * ret * a % MOD;
    a = 1ll * a * a % MOD;
    b >>= 1;
  }
  return ret;
}
void prework(int n) {
  rev[0] = 0;
  for (int i = 1 ; i < n ; ++ i)
    if (i&1) rev[i] = rev[i-1] | (n >> 1);
    else rev[i] = rev[i>>1] >> 1;
}
void dft(int* a,int n,int sgn) {
  for (int i = 0 ; i < n ; ++ i)
    if (i < rev[i]) swap(a[i],a[rev[i]]);
  int w,wp,u,v;
  for (int s = 2 ; s <= n ; s <<= 1) {
    wp = power(G,(MOD-1) / s);
    if (sgn) wp = power(wp,MOD-2);
    for (int k = 0 ; k < n ; k += s) {
      w = 1;
      for (int j = k ; j < k + (s >> 1) ; ++ j) {
	u = a[j];
	v = 1ll * w * a[j + (s >> 1)] % MOD;
	a[j] = u + v;
	if (a[j] >= MOD) a[j] -= MOD;
	a[j + (s >> 1)] = u - v + MOD;
	if (a[j + (s >> 1)] >= MOD) a[j + (s >> 1)] -= MOD;
	w = 1ll * w * wp % MOD;
      }
    }
  }
  if (sgn) {
    wp = power(n,MOD-2);
    for (int i = 0 ; i < n ; ++ i)
      a[i] = 1ll * a[i] * wp % MOD;
  }
}
int n,sz[N],jc[N],inv[N],a[N],b[N];
struct edge {
  int la,b;
} con[N];
int tot,fir[N];
void add(int from,int to) {
  con[++tot] = (edge) {fir[from],to};
  fir[from] = tot;
}
void dfs(int pos,int fa) {
  sz[pos] = 1;
  for (int i = fir[pos] ; i ; i = con[i].la) {
    if (con[i].b == fa) continue;
    dfs(con[i].b,pos);
    sz[pos] += sz[con[i].b];
    a[sz[con[i].b]] --;
  }
  a[n - sz[pos]] --;
  a[n] ++;
}
int main() {
  int x,y;
  scanf("%d",&n);
  for (int i = 1 ; i < n ; ++ i) {
    scanf("%d%d",&x,&y);
    add(x,y);
    add(y,x);
  }
  jc[0] = 1;
  for (int i = 1 ; i <= n ; ++ i)
    jc[i] = 1ll * jc[i-1] * i % MOD;
  inv[n] = power(jc[n],MOD-2);
  for (int i = n - 1 ; i >= 0 ; -- i)
    inv[i] = 1ll * inv[i+1] * (i+1) % MOD;
  dfs(1,0);
  for (int i = 0 ; i <= n ; ++ i)
    a[i] = 1ll * (a[i] + MOD) % MOD * jc[i] % MOD;
  for (int i = 0 ; i <= n ; ++ i)
    b[i] = inv[n-i];
  int l = 1;
  while (l < n + n + 2) l <<= 1;
  prework(l);
  dft(a,l,0);
  dft(b,l,0);
  for (int i = 0 ; i < l ; ++ i)
    a[i] = 1ll * a[i] * b[i] % MOD;
  dft(a,l,1);
  for (int i = 1 ; i <= n ; ++ i)
    printf("%lld\n",1ll * a[i+n] * inv[i] % MOD);
  return 0;
}
