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)

#ifdef DEBUG
#define _GLIBCXX_DEBUG
#endif
//#pragma GCC optimize("O3")
#include <bits/stdc++.h>
//THE BEST PROBLEM EVER
using namespace std;
typedef long double ld;
typedef long long ll;
int n;
const int maxN = 2 * (int)1e5 + 10;
vector < int > g[maxN];
int m;
int gr[maxN];
const int mod = 998244353;
int mult(int a, int b) {
    return (1LL * a * b) % mod;
}
int pw(int a, int b) {
    if (b == 0) return 1;
    if (b & 1) return mult(a, pw(a, b - 1));
    int res = pw(a, b / 2);
    return mult(res, res);
}
int sum(int a, int b) {
    int s = a + b;
    if (s >= mod) s -= mod;
    return s;
}
int sub(int a, int b) {
    int s = a - b;
    if (s < 0) s += mod;
    return s;
}
const int T = pw(10, 18);
int f[3][maxN];
const int LOG = 17;
void go(int v) {
    for (int i = 1; i <= n; i++) {
        g[i].clear();
    }
    cin >> m;
    for (int i = 0; i < m; i++) {
        int a, b;
        cin >> a >> b;
        if (a > b) swap(a, b);
        g[a].emplace_back(b);
    }
    for (int i = n; i >= 1; i--) {
        gr[i] = 0;
        set < int > s;
        for (int p : g[i]) {
            s.insert(gr[p]);
        }
        while (s.count(gr[i])) gr[i]++;
    }
    for (int i = 1; i <= n; i++) {
        f[v][gr[i]] = sum(f[v][gr[i]], pw(T, i));
    }
    for (int bit = 0; bit < LOG; bit++) {
        for (int i = 0; i < (1 << LOG); i++) {
            if (i & (1 << bit)) {
                int a = f[v][i ^ (1 << bit)];
                int b = f[v][i];
                f[v][i ^ (1 << bit)] = sum(a, b);
                f[v][i] = sub(a, b);
            }
        }
    }
}
int tot[maxN];
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
//    freopen("input.txt", "r", stdin);
    cin >> n;
    for (int i = 0; i < 3; i++) {
        go(i);
    }
    for (int i = 0; i < (1 << LOG); i++) {
        tot[i] = 1;
        for (int j = 0; j < 3; j++) {
            tot[i] = mult(tot[i], f[j][i]);
        }
    }
    for (int bit = 0; bit < LOG; bit++) {
        for (int i = 0; i < (1 << LOG); i++) {
            if (i & (1 << bit)) {
                int a = tot[i ^ (1 << bit)];
                int b = tot[i];
                tot[i ^ (1 << bit)] = sum(a, b);
                tot[i] = sub(a, b);
            }
        }
    }
    cout << mult(tot[0], pw((mod + 1) / 2, LOG));
    return 0;
}