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 <bits/stdc++.h>
using namespace std;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}
using Int = long long;
const char newl = '\n';


template<typename T,T MOD = 1000000007>
struct Mint{
  static constexpr T mod = MOD;
  T v;
  Mint():v(0){}
  Mint(signed v):v(v){}
  Mint(long long t){v=t%MOD;if(v<0) v+=MOD;}

  Mint pow(long long k){
    Mint res(1),tmp(v);
    while(k){
      if(k&1) res*=tmp;
      tmp*=tmp;
      k>>=1;
    }
    return res;
  }

  static Mint add_identity(){return Mint(0);}
  static Mint mul_identity(){return Mint(1);}

  Mint inv(){return pow(MOD-2);}

  Mint& operator+=(Mint a){v+=a.v;if(v>=MOD)v-=MOD;return *this;}
  Mint& operator-=(Mint a){v+=MOD-a.v;if(v>=MOD)v-=MOD;return *this;}
  Mint& operator*=(Mint a){v=1LL*v*a.v%MOD;return *this;}
  Mint& operator/=(Mint a){return (*this)*=a.inv();}

  Mint operator+(Mint a) const{return Mint(v)+=a;}
  Mint operator-(Mint a) const{return Mint(v)-=a;}
  Mint operator*(Mint a) const{return Mint(v)*=a;}
  Mint operator/(Mint a) const{return Mint(v)/=a;}

  Mint operator-() const{return v?Mint(MOD-v):Mint(v);}

  bool operator==(const Mint a)const{return v==a.v;}
  bool operator!=(const Mint a)const{return v!=a.v;}
  bool operator <(const Mint a)const{return v <a.v;}

  static Mint comb(long long n,int k){
    Mint num(1),dom(1);
    for(int i=0;i<k;i++){
      num*=Mint(n-i);
      dom*=Mint(i+1);
    }
    return num/dom;
  }
};
template<typename T,T MOD> constexpr T Mint<T, MOD>::mod;
template<typename T,T MOD>
ostream& operator<<(ostream &os,Mint<T, MOD> m){os<<m.v;return os;}

//INSERT ABOVE HERE
signed main(){
  cin.tie(0);
  ios::sync_with_stdio(0);

  int n;
  cin>>n;

  vector<vector<int>> G[3];
  for(int t=0;t<3;t++){
    G[t].resize(n);
    int m;
    cin>>m;
    for(int i=0;i<m;i++){
      int a,b;
      cin>>a>>b;
      a--;b--;
      if(a>b) swap(a,b);
      G[t][a].emplace_back(b);
    }
  }

  const int MAX = 1010;
  using M = Mint<int, 998244353>;
  M base=M(10).pow(18);

  vector<M> cnt[3];
  for(int t=0;t<3;t++){
    vector<int> dp(n);
    dp.resize(n,0);
    for(int i=n-1;i>=0;i--){
      set<int> used;
      for(int j:G[t][i])
        used.emplace(dp[j]);
      while(used.count(dp[i])) dp[i]++;
    }
    cnt[t].resize(MAX,0);
    for(int i=0;i<n;i++) cnt[t][dp[i]]+=base.pow(i+1);
  }

  M ans{0};
  for(int x=0;x<MAX;x++)
    for(int y=0;y<MAX;y++)
      ans+=cnt[0][x]*cnt[1][y]*cnt[2][x^y];

  cout<<ans<<newl;
  return 0;
}