#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define fi first
#define se second
#define mp make_pair
#define eb emplace_back
#define ALL(x) (x).begin(), (x).end()
#define FOR(i, a, b) for(int i = (a); i < (b); ++i)
#define FORU(i, a, b) for(int i = (a); i <= (b); ++i)
#define FORD(i, a, b) for(int i = (a); i >= (b); --i)
#define IOS cin.tie(0)->sync_with_stdio(false);
#define PROB "CEOI19_amusementpark"
void Fi(){
if(fopen(PROB".inp", "r")){
freopen(PROB".inp", "r", stdin);
freopen(PROB".out", "w", stdout);
}
}
const int N = 18;
const int MOD = 998244353;
int n, m;
int edge[N * N];
int f[1 << N][N + 1]; /// number of not connected graph
int dp[1 << N][N + 1]; /// total number of graph
bool noEdge(const int &mask){
FOR(i, 0, m) if((mask & edge[i]) == edge[i]) return false;
return true;
}
void addto(int &a, const int &b){
a = (a + b >= MOD ? a + b - MOD : a + b);
}
void subto(int &a, const int &b){
a = (a - b < 0 ? a - b + MOD : a - b);
}
int mul(const int &a, const int &b){
return (1LL * a * b) % MOD;
}
int main(){
IOS;
Fi();
cin >> n >> m;
FOR(i, 0, m){
int u, v; cin >> u >> v;
--u, --v;
edge[i] = (1 << u) | (1 << v);
}
FOR(mask, 0, 1 << n) if(noEdge(mask)){
int nbit = __builtin_popcount(mask);
f[mask][nbit] = ((nbit & 1) ? 1 : MOD - 1);
} /// init the no edge mask
for(int i = 1; i < (1 << n); i <<= 1){ /// can remove bit 0...(log2(i))
for(int j = 0; j < (1 << n); j += (i << 1)){
FOR(k, 0, i) FORU(p, 0, n) addto(f[j | i | k][p], f[j | k][p]);
}
}
/// calculate f[mask][i]: number of distinct subset of [mask] have [i] bit and not connected
/// use -1 with odd and even +1 to use inclusive and exclusive for calculate dp later
FOR(mask, 0, 1 << n){
dp[mask][0] = 1;
FORU(i, 1, n) FOR(j, 0, i){
addto(dp[mask][i], mul(dp[mask][j], f[mask][i - j]));
}
}
/// calculate dp[mask][i]: total number of distinct subset of [mask] have [i] bit
for(int i = 1; i < (1 << n); i <<= 1){
for(int j = 0; j < (1 << n); j += (i << 1)){
FOR(k, 0, i) FORU(p, 0, n) subto(dp[j | i | k][p], dp[j | k][p]);
}
}
/// subtract to get the answer for each distinct set
int ans = dp[(1 << n) - 1][n];
ans = mul(ans, m); /// there are m edges
ans = mul(ans, (MOD + 1) / 2); /// devide by 2 (the not reverse edges)
cout << ans;
return 0;
}
/// Idea stolen
