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 <algorithm>
#include <cstring>
#include <vector>
 
#define Rep(i, n) for (int i = 1; i <= n; i ++)
#define Rep0(i, n) for (int i = 0; i <= n; i ++)
#define RepG(i, x) for (int i = head[x]; i; i = edge[i].next)
#define v edge[i].to
#define mp(a, b) make_pair(a, b)
 
using namespace std;
 
typedef long long LL;
const int N = 2000010;
const int mod = 924844033;
const int G = 5;
 
LL qpow(LL a, LL b)
{
    LL ret = 1, base = a;
    while (b) {
        if (b & 1) ret = ret * base % mod;
        base = base * base % mod, b >>= 1;
    }
    return ret;
}
 
void NTT(LL *a, int len, int f)
{
    for (int i = 0, j = 0; i < len; i ++) {
        if (i < j) swap(a[i], a[j]);
        for (int t = len >> 1; (j ^= t) < t; t >>= 1);
    }
    for (int i = 1; i < len; i <<= 1) {
        LL wn = f ? qpow(G, (mod - 1) / (i << 1)) : qpow(G, (mod - 1) - (mod - 1) / (i << 1));
        for (int j = 0; j < len; j += (i << 1)){
            LL w = 1;
            for (int k = 0; k < i; k ++, w = w * wn % mod){
                LL x = a[j + k], y = w * a[j + k + i] % mod;
                a[j + k] = (x + y) % mod;
                a[j + k + i] = (x - y + mod) % mod;
            }
        }
    }
    if (!f) for (int i = 0; i < len; i ++) a[i] = a[i] * qpow(len, mod - 2) % mod;
}
 
LL p[N], invp[N];

struct Edge{ int to, next;} edge[N * 2];
int n, head[N], num;
void add_edge(int a, int b) { edge[++ num] = (Edge){b, head[a]}, head[a] = num;}

LL a[N], b[N];

int dfs(int x, int fa)
{
	int sz = 1;
	RepG(i, x) if (v != fa) {
		int tmp = dfs(v, x);
		sz += tmp, a[n - tmp] ++;
	}
	a[sz] ++;
	return sz;
}

int main()
{
	scanf("%d", &n);
	Rep(i, n - 1) {
		int a, b;
		scanf("%d%d", &a, &b);
		add_edge(a, b); add_edge(b, a);
	}
	dfs(1, 0);
	p[0] = invp[0] = 1;
	Rep(i, n) p[i] = p[i - 1] * i % mod, invp[i] = qpow(p[i], mod - 2);
	
	Rep0(i, n){
		a[i] = a[i] * p[n - i] % mod;
		b[i] = invp[i];
	}
	
	int len = 1;
	while (len <= n << 1) len <<= 1;

	NTT(a, len, 1); NTT(b, len, 1);
	Rep0(i, len) a[i] = a[i] * b[i] % mod;
	NTT(a, len, 0);
	Rep(i, n){
		LL tmp = p[n] * invp[i] % mod * invp[n - i] % mod;
		printf("%lld\n", (n * tmp % mod - a[n - i] * invp[i] % mod + mod) % mod);
	}
	
	return 0;
}