//vanitas vanitatum et omnia vanitas
#include<bits/stdc++.h>
#define fi first
#define se second
#define eb emplace_back
#define mp make_pair
using namespace std;
typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef __int128 i128; 
template<typename T,typename U>
T ceil(T x, U y) {return (x>0?(x+y-1)/y:x/y);}
template<typename T,typename U>
T floor(T x, U y) {return (x>0?x/y:(x-y+1)/y);}
template<class T,class S>
bool chmax(T &a,const S b) {return (a<b?a=b,1:0);}      
template<class T,class S>
bool chmin(T &a,const S b) {return (a>b?a=b,1:0);}
int popcnt(int x) {return __builtin_popcount(x);}
int popcnt(ll x)  {return __builtin_popcountll(x);}
int topbit(int x) {return (x==0?-1:31-__builtin_clz(x));}
int topbit(ll x)  {return (x==0?-1:63-__builtin_clzll(x));}
int lowbit(int x) {return (x==0?-1:__builtin_ctz(x));}
int lowbit(ll x)  {return (x==0?-1:__builtin_ctzll(x));}

#define int long long
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
typedef pair<int,int> pii; 
typedef vector<int> vi;  
typedef vector<pii> vp; 
typedef tuple<int,int,int> tiii;
int read() {
  int x=0,w=1; char c=getchar(); 
  while(!isdigit(c)) {if(c=='-') w=-1; c=getchar();}
  while(isdigit(c)) {x=x*10+(c-'0'); c=getchar();}
  return x*w;
}

const int N=1005,mod=998244353;
int n,m,p[N],ans;
bool e[N][N];

void solve(vi &s,int pp) {
  //cout<<"SOLVE: ("<<pp<<") ";
  //sort(s.begin(),s.end());
  //for(int x:s) cout<<x<<" "; puts("");
  //for(int x:s) {
  //  for(int y:s) cout<<e[x][y];
  //    puts("");
  //}
  if(s.size()==0) {ans=(ans+pp*(!e[0][0]))%mod; return;}
  else if(s.size()==1) {
    e[0][s[0]]^=e[s[0]][s[0]];
    ans=(ans+pp*(1+(!e[0][s[0]])))%mod;
    return;
  }
  for(int x:s) if(e[x][x]) e[0][x]^=1, e[x][0]^=1, e[x][x]=0;
  int u=s[0], v=0, n=s.size();
  for(int x:s) if(e[u][x]) {v=x; break;}
  if(!v) {
    if(e[0][u]) {ans=(ans+p[n-1]*pp)%mod; return;}
    vi t; for(int x:s) if(x!=u) t.eb(x);
    solve(t,pp*2%mod); return;
  }
  ans=(ans+p[n-2]*pp)%mod; //cout<<p[n-2]*pp<<endl;
  static int a[N],b[N];
  for(int x:s) a[x]=e[u][x], b[x]=e[v][x];
  vi t; for(int x:s) if(x!=u&&x!=v) t.eb(x);
  for(int x:t) for(int y:t) if(x<=y) {
    //cout<<x<<" "<<y<<" "<<b[x]<<" "<<a[y]<<endl;
    if(x==y) e[x][y]^=(a[x]&&b[x]);
    else {
      if(a[x]&&b[y]) e[x][y]^=1, e[y][x]^=1;
      if(b[x]&&a[y]) e[x][y]^=1, e[y][x]^=1;
    }
  }
  e[0][0]^=e[u][0]&&e[v][0];
  solve(t,pp*2%mod);
}

signed main() {
  n=read(), m=read();
  p[0]=1; rep(i,1,n) p[i]=p[i-1]*2%mod;
  rep(i,1,m) {
    int u=read(), v=read();
    e[u][v]^=1, e[v][u]^=1;
  }
  vi g; rep(i,1,n) g.eb(i);
  solve(g,1); printf("%lld\n",ans);
  return 0;
}