若连通块是一棵树，考虑钦定 \(k\) 个点为奇度点，方案数为 \(\binom{n}{k}\)。对于叶子，如果它是奇度点，那么连向它父亲的边要保留，否则不保留。这样自底向上考虑，任意一条边的保留情况都可以唯一确定，所以最后方案数就是 \(\binom{n}{k}\)。
若连通块存在环，仍然钦定 \(k\) 个点为奇度点。考虑随便拎出一棵生成树，非树边选可不选；确定了非树边的状态，就和树的情况一样了。设连通块边数为 \(m\)，最后方案数就是 \(2^{m-n+1} \binom{n}{k}\)。
多个连通块的情况，考虑背包 dp，设 \(f_{i,j}\) 为前 \(i\) 个连通块，选了 \(j\) 个奇度点的方案数。转移枚举第 \(i\) 个连通块钦定 \(k\) 个奇度点（\(k\) 为偶数）即可。
时间复杂度 \(O(n^2)\)。
code
// Problem: D - Odd Degree
// Contest: AtCoder - AtCoder Regular Contest 115
// URL: https://atcoder.jp/contests/arc115/tasks/arc115_d
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<int, int> pii;
const int maxn = 5050;
const int N = 5000;
const ll mod = 998244353;
inline ll qpow(ll b, ll p) {
ll res = 1;
while (p) {
if (p & 1) {
res = res * b % mod;
}
b = b * b % mod;
p >>= 1;
}
return res;
}
int n, m, a[maxn], sz[maxn], fa[maxn];
ll fac[maxn], ifac[maxn], pw[maxn], f[maxn][maxn];
pii E[maxn];
int find(int x) {
return fa[x] == x ? x : fa[x] = find(fa[x]);
}
inline void merge(int x, int y) {
x = find(x);
y = find(y);
if (x != y) {
fa[x] = y;
sz[y] += sz[x];
}
}
void init() {
pw[0] = fac[0] = 1;
for (int i = 1; i <= N; ++i) {
pw[i] = pw[i - 1] * 2 % mod;
fac[i] = fac[i - 1] * i % mod;
}
ifac[N] = qpow(fac[N], mod - 2);
for (int i = N - 1; ~i; --i) {
ifac[i] = ifac[i + 1] * (i + 1) % mod;
}
}
inline ll C(ll n, ll m) {
if (n < m || n < 0 || m < 0) {
return 0;
} else {
return fac[n] * ifac[m] % mod * ifac[n - m] % mod;
}
}
void solve() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++i) {
fa[i] = i;
sz[i] = 1;
}
for (int i = 1; i <= m; ++i) {
scanf("%d%d", &E[i].fst, &E[i].scd);
merge(E[i].fst, E[i].scd);
}
for (int i = 1; i <= m; ++i) {
++a[find(E[i].fst)];
}
f[0][0] = 1;
int t = 0;
for (int i = 1; i <= n; ++i) {
if (fa[i] != i) {
continue;
}
++t;
for (int j = 0; j <= n; ++j) {
for (int k = 0; k <= min(sz[i], j); k += 2) {
f[t][j] = (f[t][j] + f[t - 1][j - k] * C(sz[i], k) % mod * pw[a[i] - (sz[i] - 1)] % mod) % mod;
}
}
}
for (int i = 0; i <= n; ++i) {
printf("%lld\n", f[t][i]);
}
}
int main() {
init();
int T = 1;
// scanf("%d", &T);
while (T--) {
solve();
}
return 0;
}