For this Question: You are given a simple undirected graph consisting of N nodes and M edges. Count the number of distinct ways of removing a (possibly empty) subset of edges such that the <b>remaining graph on N nodes remains edge biconnected</b>.
<ul>
<li> A graph is said to be <b>edge biconnected</b> if, for every pair of distinct nodes, there exist two edge-disjoint paths between them. Two paths are edge-disjoint if they do not share an edge.
</ul>
The number of ways may be large, so print the answer modulo 998244353.The first line of the input contains two integers, N and M, the number of nodes in the graph and the number of edges in the graph.
M lines follow, each containing 2 integers, u<sub>i</sub> and v<sub>i</sub> denoting an edge in the graph. 

<b>Constraints</b>
3 <= N <= 17
1 <= M <= (N*(N-1)) / 2
1 <= u<sub>i</sub>, v<sub>i</sub> <= N for all 1 <= i <= M
It is guaranteed that there are no self-loops and multiple edges in the graph.
Print a single integer, the required number of ways modulo 998244353.Sample Input 1:
4 5
1 2
2 3
3 4
4 1
4 2

Sample Output 1:
2

<b>Explanation:</b>
For the first sample input, the two ways are to remove either the empty set of edges or to remove the edge indexed 5.

Sample Input 2:
4 6
1 2
2 3
3 4
4 1
1 3
2 4


Sample Output 2:
10

<b>Explanation:</b>
For the second sample, it can be shown that there are 6 ways to remove 1 edge, 3 ways to remove 2 edges, and 1 way to remove 0 edges., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define debug(...) fprintf (stderr, __VA_ARGS__)

#define fi first
#define se second
#define Lep(i, l, r) for (int i = (l), i##_end = (r); i < i##_end; ++ i)
#define lep(i, l, r) for (int i = (l), i##_end = (r); i <= i##_end; ++ i)
#define rep(i, r, l) for (int i = (r), i##_end = (l); i >= i##_end; -- i)

char __c; bool __f; template <class T> inline void IN (T & x) {
	__f = 0, x = 0; while (__c = getchar (), ! isdigit (__c)) if (__c == '-') __f = 1;
	while (isdigit (__c)) x = x * 10 + __c - '0', __c = getchar (); if (__f) x = -x;
}

template <class T> inline void chkmin (T & x, T y) { if (x > y) x = y; }
template <class T> inline void chkmax (T & x, T y) { if (x < y) x = y; }

const int mod = 998244353;

inline int mul (int x, int y) { return 1ll * x * y % mod; }
inline void pls (int &x, int y) { x += y; if (x >= mod) x -= mod; }
inline int add (int x, int y) { x += y; if (x >= mod) x -= mod; return x; }

inline int modpow (int x, int y, int r = 1) {
	for ( ; y; y >>= 1, x = mul (x, x)) if (y & 1) r = mul (r, x);
	return r;
}

const int N = 20;
const int NS = (1 << 18) | 5;

int inv[N];
inline void init (int L = N - 2) {
	inv[1] = 1;
	lep (i, 2, L) inv[i] = mod - mul (inv[mod % i], mod / i);
}

inline void fmt (int f[NS][N], int n) {
	Lep (p, 0, n) for (int k = 0; k < (1 << n); k += (1 << p + 1))
		Lep (l, k, k | (1 << p)) {
			lep (d, 1, n) pls (f[l | (1 << p)][d], f[l][d]);
		}
}
inline void ifmt (int f[NS][N], int n) {
	Lep (p, 0, n) for (int k = 0; k < (1 << n); k += (1 << p + 1))
		Lep (l, k, k | (1 << p)) {
			lep (d, 1, n) pls (f[l | (1 << p)][d], mod - f[l][d]);
		}
}

inline void ln (int *f, int n) {
	static int g[N];
	lep (i, 0, n - 1) {
		g[i] = mul (f[i + 1], i + 1);
		lep (j, 0, i - 1) pls (g[i], mod - mul (g[j], f[i - j]));
	}
	lep (i, 0, n - 1) f[i + 1] = mul (g[i], inv[i + 1]);
}
inline void exp (int *f, int n) {
	static int g[N];
	lep (i, 0, n - 1) g[i] = mul (f[i + 1], i + 1), f[i + 1] = 0;
	lep (i, 0, n - 1) {
		lep (j, 0, i - 1) pls (f[i + 1], mul (g[j], f[i - j]));
		f[i + 1] = mul (add (g[i], f[i + 1]), inv[i + 1]);
	}
}

int n, m, G[N];
int sf[NS][N];

signed main () {
	init ();

	IN (n), IN (m);
	for (int u, v; m; -- m) {
		IN (u), IN (v), -- u, -- v;
		G[u] |= 1 << v, G[v] |= 1 << u;
	}
	if (n == 1) return puts ("1"), 0;

	Lep (S, 1, 1 << n) {
		int cnt = 0;
		Lep (i, 0, n) if ((S >> i) & 1) cnt += __builtin_popcount (G[i] & S);
		sf[S][__builtin_popcount (S)] = modpow (2, cnt / 2);
	}

	fmt (sf, n);
	Lep (S, 1, 1 << n) ln (sf[S], n);

	Lep (i, 0, n) {
		for (int k = 0; k < (1 << n); k += (1 << i + 1))
			Lep (l, k, k | (1 << i)) {
				lep (d, 1, n) pls (sf[l | (1 << i)][d], mod - sf[l][d]);
				ln (sf[l | (1 << i)] + 1, n);
				lep (d, 1, n) pls (sf[l | (1 << i)][d], sf[l][d]);
			}
	}

	ifmt (sf, n);
	Lep (S, 1, 1 << n) if (__builtin_popcount (S) <= 2) sf[S][__builtin_popcount (S)] = 0;
	fmt (sf, n);

	Lep (i, 0, n) {
		for (int k = 0; k < (1 << n); k += (1 << i + 1))
			Lep (l, k, k | (1 << i)) {
				lep (d, 1, n) pls (sf[l | (1 << i)][d], mod - sf[l][d]);
				exp (sf[l | (1 << i)] + 1, n);
				lep (d, 1, n) pls (sf[l | (1 << i)][d], sf[l][d]);
			}
	}

	ifmt (sf, n);
	printf ("%d\n", sf[(1 << n) - 1][n]);
	return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not?