Problem:
Score : 900 points

Problem Statement

Given are simple undirected graphs X, Y, Z, with N vertices each and M_1, M_2, M_3 edges, respectively.
The vertices in X, Y, Z are respectively called x_1, x_2, \dots, x_N, y_1, y_2, \dots, y_N, z_1, z_2, \dots, z_N.
The edges in X, Y, Z are respectively (x_{a_i}, x_{b_i}), (y_{c_i}, y_{d_i}), (z_{e_i}, z_{f_i}).

Based on X, Y, Z, we will build another undirected graph W with N^3 vertices.
There are N^3 ways to choose a vertex from each of the graphs X, Y, Z. Each of these choices corresponds to the vertices in W one-to-one. Let (x_i, y_j, z_k) denote the vertex in W corresponding to the choice of x_i, y_j, z_k.

We will span edges in W as follows:

For each edge (x_u, x_v) in X and each w, l, span an edge between (x_u, y_w, z_l) and (x_v, y_w, z_l).

For each edge (y_u, y_v) in Y and each w, l, span an edge between (x_w, y_u, z_l) and (x_w, y_v, z_l).

For each edge (z_u, z_v) in Z and each w, l, span an edge between (x_w, y_l, z_u) and (x_w, y_l, z_v).

Then, let the weight of the vertex (x_i, y_j, z_k) in W be 1,000,000,000,000,000,000^{(i +j + k)} = 10^{18(i + j + k)}. Find the maximum possible total weight of the vertices in an independent set in W, and print that total weight modulo 998,244,353.

Constraints

2 \leq N \leq 100,000

1 \leq M_1, M_2, M_3 \leq 100,000

1 \leq a_i, b_i, c_i, d_i, e_i, f_i \leq N

X, Y, and Z are simple, that is, they have no self-loops and no multiple edges.

Input

Input is given from Standard Input in the following format:

N
M_1
a_1 b_1
a_2 b_2
\vdots
a_{M_1} b_{M_1}
M_2
c_1 d_1
c_2 d_2
\vdots
c_{M_2} d_{M_2}
M_3
e_1 f_1
e_2 f_2
\vdots
e_{M_3} f_{M_3}

Output

Print the maximum possible total weight of an independent set in W, modulo 998,244,353.

Sample Input 1

2
1
1 2
1
1 2
1
1 2

Sample Output 1

46494701

The maximum possible total weight of an independent set is that of the set (x_2, y_1, z_1), (x_1, y_2, z_1), (x_1, y_1, z_2), (x_2, y_2, z_2). The output should be (3 \times 10^{72} + 10^{108}) modulo 998,244,353, which is 46,494,701.

Sample Input 2

3
3
1 3
1 2
3 2
2
2 1
2 3
1
2 1

Sample Output 2

883188316

Sample Input 3

100000
1
1 2
1
99999 100000
1
1 100000

Sample Output 3

318525248

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<vector>
#include<set>
#include<map>
#include<queue>
#include<string>
#include<algorithm>
#include<iostream>
#include<bitset>
#include<functional>
#include<numeric>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cassert>
#include<cmath>
#include<random>
#include<ctime>
#include<complex>
using namespace std;
typedef long long LL;
mt19937 gene(233);
typedef complex<double> Complex;
#define fi first
#define se second
#define ins insert
#define pb push_back
inline char GET_CHAR(){
	const int maxn = 131072;
	static char buf[maxn],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,maxn,stdin),p1==p2)?EOF:*p1++;
}
inline int getInt() {
	int res(0);
	char c = getchar();
	while(c < '0') c = getchar();
	while(c >= '0') {
		res = res * 10 + (c - '0');
		c = getchar();
	}
	return res;
}

inline int fastpo(int x, int n, int mod) {
	int res(1);
	while(n) {
		if(n & 1) {
			res = res * (LL)x % mod;
		}
		x = x * (LL) x % mod;
		n /= 2;
	}
	return res;
}

inline string itoa(int x, int width = 0) {
  string res;
  if(x == 0) res.push_back('0');
  while(x) {
    res.push_back('0' + x % 10);
    x /= 10;
  }
  while((int)res.size() < width) res.push_back('0');
  reverse(res.begin(), res.end());
  return res;
}

const int N = 131072;
const int LOG = 17;
const int mod = 998244353;
const int inf = 1e9 + 7;
int n, m;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
vector<int> e[N];
int vst[N], par[N];
struct M {
	LL a;
	M operator + (const M & b) const {
		M res;
		res.a = a + b.a;
		if(res.a >= mod) res.a -= mod;
		return res;
	}
	M operator - (const M & b) const {
		M res;
		res.a = a - b.a;
		if(res.a < 0) res.a += mod;
		return res;
	}
	M operator * (const M & b) const {
		return M{a * b.a % mod};
	}
};
M s[3][N];
M pw[N];
set<int> st[N];
int main() {
	int n;
	scanf("%d", &n);
	pw[0] = M{1};
	int bs = 1000000000000000000ll % mod;
	for(int i = 1; i <= n; i++) pw[i] = pw[i - 1] * M{bs};
	for(int d = 0; d < 3; d++) {
		int m;
		scanf("%d", &m);
		for(int i = 1; i <= n; i++) e[i].clear();
		for(int i = 1; i <= m; i++) {
			int x, y;
			scanf("%d%d", &x, &y);
			if(y < x) swap(x, y);
//			e[x].push_back(y);
			e[y].push_back(x);
		}
		for(int i = 1; i <= n; i++) {
			st[i].clear();
		}
		for(int i(n); i >= 1; i--) {
			int val = 0;
			while(st[i].count(val)) val++;
			//cout << val << endl;
			s[d][val] = s[d][val] + pw[i];
			for(int y : e[i]) {
				st[y].insert(val);
			}
		}
		for(int i = 0; i < LOG; i++) {
			for(int j = 0; j < N; j++) {
				if((j >> i) & 1) {
					M a = s[d][j], b = s[d][j - (1 << i)];
					s[d][j] = a + b;
					s[d][j - (1 << i)] = b - a;
				}
			}
		}
	}
	for(int i = 0; i < N; i++) {
		s[0][i] = s[1][i] * s[0][i] * s[2][i];
	}
	//cout << s[0][0].a << ' ' << s[0][1].a << endl;
	for(int i = LOG - 1; i >= 0; i--) {
		for(int j = 0; j < N; j++) {
			if((j >> i) & 1) {
				M a = s[0][j], b = s[0][j - (1 << i)];
				s[0][j] = (a - b) * M{(mod + 1) / 2};
				s[0][j - (1 << i)] = (a + b) * M{(mod + 1) / 2};
			}
		}
	}
	cout << s[0][0].a << endl;
}