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 <cstdio>
#include <cstring>
#include <algorithm>
#define MOD 924844033
 
using namespace std;
 
typedef long long ll;
 
ll pow_mod(ll x,int k) {
ll ans=1;
while (k) {
if (k&1) ans=ans*x%MOD;
x=x*x%MOD;
k>>=1;
}
return ans;
}
 
ll a[1<<19],b[1<<19];
 
void rev(ll *p,int len) {
int j=len>>1;
for(int i=1;i<len-1;i++) {
if (i<j) swap(p[i],p[j]);
int k=len>>1;
while (j>=k) {
j-=k;
k>>=1;
}
if (j<k) j+=k;
}
}
 
void ntt(ll *p,int len,int check) {
rev(p,len);
for(int i=2;i<=len;i<<=1) {
ll wn=pow_mod(5,(MOD-1)/i);
for(int j=0;j<len;j+=i) {
ll w=1;
for(int k=j;k<j+(i>>1);k++) {
ll u=p[k];
ll v=w*p[k+(i>>1)]%MOD;
p[k]=(u+v)%MOD;
p[k+(i>>1)]=(u-v+MOD)%MOD;
w=w*wn%MOD;
}
}
}
if (check==-1) {
reverse(p+1,p+len);
ll nev=pow_mod(len,MOD-2);
for(int i=0;i<len;i++) p[i]=p[i]*nev%MOD;
}
}
 
struct Edge {
int t,next;
Edge() {}
Edge(int a,int b):t(a),next(b) {}
};
 
Edge e[400005];
int head[200005],num[200005],n;
 
int size[200005];
 
void dfs(int x,int fa) {
size[x]=1;
for(int i=head[x];i;i=e[i].next)
if (e[i].t!=fa) {
int u=e[i].t;
dfs(u,x);
size[x]+=size[u];
num[n-size[u]]++;
num[size[u]]++;
}
}
 
ll facd[200005],facv[200005];
 
void pre(int n) {
facd[0]=1;
for(int i=1;i<=n;i++) facd[i]=facd[i-1]*i%MOD;
facv[n]=pow_mod(facd[n],MOD-2);
for(int i=n-1;i>=0;i--) facv[i]=facv[i+1]*(i+1)%MOD;
}
 
int main() {
scanf("%d",&n);
pre(n);
for(int i=1;i<n;i++) {
int x,y;
scanf("%d%d",&x,&y);
e[2*i-1]=Edge(y,head[x]);
head[x]=2*i-1;
e[2*i]=Edge(x,head[y]);
head[y]=2*i;
}
dfs(1,0);
int len=1;
while (len<((n+1)<<1)) len<<=1;
for(int i=0;i<n;i++) {
a[i]=num[i]*facd[n-i]%MOD;
b[i]=facv[i];
}
ntt(a,len,1);ntt(b,len,1);
for(int i=0;i<len;i++) a[i]=a[i]*b[i]%MOD;
ntt(a,len,-1);
for(int i=1;i<=n;i++) {
ll s1=facd[n]*facv[i]%MOD*facv[n-i]%MOD*n%MOD;
ll s2=a[n-i]*facv[i]%MOD;
printf("%lld\n",(s1-s2+MOD)%MOD);
}
return 0;
}
