Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=(a),i##ed=(b);i<=i##ed;i++)
#define per(i,a,b) for(int i=(a),i##ed=(b);i>=i##ed;i--)
#define LNK(i,a) for(int i=hed[a];i;i=nxt[i])
using namespace std;
typedef long long ll;
const ll mod=924844033;
const int N=200010;
int n,num[N],sz[N];
int cnt=0,hed[N],to[N+N],nxt[N+N];
ll fa[N],fb[N];
ll a[N*4],b[N*4]; int lmt,l,r[N*4];

ll ksm(ll x,ll y) { ll s=1;for(;y;y/=2,x=x*x%mod) if(y%2) s=s*x%mod;return s; }
inline void add(int x,int y) { to[++cnt]=y,nxt[cnt]=hed[x],hed[x]=cnt; }
void dfs(int u,int ff) {
	sz[u]=1;
	LNK(i,u) if(to[i]^ff)
		dfs(to[i],u),sz[u]+=sz[to[i]],--num[sz[to[i]]],--num[n-sz[to[i]]];
}
void ntt(ll *A,int opt) {
	rep(i,0,lmt-1) if(i<r[i]) swap(A[i],A[r[i]]);
	for(int mid=1;mid<lmt;mid<<=1) {
		ll Wn=ksm(opt==1? 5:554906420,(mod-1)/(mid<<1));
		for(int R=mid<<1,j=0;j<lmt;j+=R) {
			ll w=1;
			for(int k=0;k<mid;k++,w=w*Wn%mod) {
				ll x=A[j+k],y=w*A[j+mid+k]%mod;
				A[j+k]=(x+y)%mod,A[j+mid+k]=(x-y+mod)%mod;
			}
		}
	}
	if(opt==-1) {
		ll inv=ksm(lmt,mod-2); rep(i,0,lmt-1) a[i]=a[i]*inv%mod;
	}
}
int main() {
	scanf("%d",&n);
	rep(i,2,n) { int x,y;scanf("%d%d",&x,&y),add(x,y),add(y,x); }
	rep(i,1,n) num[i]=mod; dfs(1,0),num[n]=n; rep(i,1,n) num[i]%=mod;
//	rep(i,1,n) printf("%d ",num[i]); puts("");
	fa[0]=fa[1]=fb[0]=fb[1]=1;
	rep(i,2,n) fa[i]=fa[i-1]*i%mod;
	rep(i,2,n) fb[i]=(mod-mod/i)*fb[mod%i]%mod;
	rep(i,2,n) fb[i]=fb[i-1]*fb[i]%mod;
	rep(i,1,n) a[n-i]=fa[i]*num[i]%mod,b[i-1]=fb[i-1];
//	rep(i,0,n-1) printf("%lld ",a[i]); puts("");
//	rep(i,0,n-1) printf("%lld ",b[i]); puts("");
	for(lmt=1,l=0;lmt<=n+n;lmt<<=1,++l);
	rep(i,0,lmt-1) r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
	ntt(a,1),ntt(b,1);
	rep(i,0,lmt-1) a[i]=a[i]*b[i]%mod;
	ntt(a,-1);
//	rep(i,0,n+n-2) printf("%lld ",a[i]); puts("");
	rep(i,1,n) printf("%lld\n",(a[n-i]*fb[i]%mod+mod)%mod);
	return 0;
}