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 pb push_back
#define Inv(x) ((ll)fac[x - 1] * inv[x] % mod) 

using namespace std;

typedef long long ll;

const int mod = 924844033, N = 8e5 + 10;
int a[N], b[N], rev[N];
int n, cnt[N], size[N];
int fac[N], inv[N];
vector<int> G[N];

int C(int n, int m) {
	if(n < m) return 0;
	return (ll)fac[n] * inv[m] % mod * inv[n - m] % mod;
}

int qpow(int a, int x) {
	int ret = 1;
	while(x) {
		if(x & 1) ret = (ll)ret * a % mod;
		x >>= 1, a = (ll)a * a % mod;
	}
	return ret;
}

void dfs(int x) {
	size[x] = 1;
	for(auto v : G[x])
		if(!size[v]) {
			dfs(v);
			size[x] += size[v];
		}
	if(size[x] - n)
		++ cnt[size[x]], ++ cnt[n - size[x]];
}

void NTT(int n, int *a, int fh) {
	for(int i = 0; i < n; ++ i)
		if(i < rev[i]) 
			swap(a[i], a[rev[i]]);
	for(int limit = 2; limit <= n; limit <<= 1) {
		int Wn = qpow(fh == 1 ? 5 : Inv(5), (mod - 1) / limit), W = 1;
		for(int j = 0; j < n; j += limit, W = 1)
			for(int i = j; i < j + (limit >> 1); ++ i, W = (ll)W * Wn % mod) {
				int a1 = a[i], a2 = (ll)a[i + (limit >> 1)] * W % mod;
				a[i] = (a1 + a2) % mod, a[i + (limit >> 1)] = (a1 - a2 + mod) % mod;
			}
	}
	if(fh == -1) 
		for(int i = 0; i < n; ++ i)
			a[i] = ll(a[i]) * Inv(n) % mod;
}

int main() {
//	freopen("AGC005F.in", "r", stdin);
//	freopen("AGC005F.out", "w", stdout);
	int x, y, limit = 1, k = 0;
	scanf("%d", &n);
	fac[0] = inv[0] = 1;
	for(int i = 1; i <= N - 5; ++ i) 
		fac[i] = (ll)fac[i - 1] * i % mod;
	inv[N - 5] = qpow(fac[N - 5], mod - 2);
	for(int i = N - 5; i >= 1; -- i) 
		inv[i - 1] = (ll)inv[i] * i % mod;
	for(int i = 1; i <= n; ++ i) {
		scanf("%d%d", &x, &y);
		G[x].pb(y), G[y].pb(x);	
	}
	dfs(1);
	for(int i = 0; i <= n; ++ i) 
		a[i] = (ll)cnt[i] * fac[i] % mod;
	for(int i = 0; i <= n; ++ i)
		b[i] = inv[i];
	reverse(b, b + n + 1);
	while(limit <= (n << 1)) 
		limit <<= 1, ++ k;
	for(int i = 0; i < limit; ++ i)
		rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (k - 1));
	NTT(limit, a, 1), NTT(limit, b, 1);
	for(int i = 0; i < limit; ++ i)
		a[i] = (ll)a[i] * b[i] % mod;
	NTT(limit, a, -1);
	//for(int i = 0; i < limit; ++ i)
	//	printf("%d ", a[i]);
	for(int i = (n << 1) - 1; i >= n; -- i) {
		int k = (n << 1) - i;
		printf("%d ", (((ll)n * C(n, k) - (ll)a[k + n] * inv[k]) % mod + mod) % mod);
	}
	return 0;
}