Problem: 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.

Examples

Input

2
1
1 2
1
1 2
1
1 2


Output

46494701


Input

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


Output

883188316


Input

100000
1
1 2
1
99999 100000
1
1 100000


Output

318525248

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <cassert>
#include <cmath>
#include <algorithm>

using ll = long long;

#define MIN(a, b) (((a) < (b)) ? (a) : (b))
#define MAX(a, b) (((a) < (b)) ? (b) : (a))

int const nmax = 100000;
int const rad = 500;
int const modulo = 998244353;
std::vector<int> g[3][1 + nmax];
int nimber[3][1 + nmax];

int lgpow(int a, int b) {
  if(b == 0)
    return 1;
  else if(b == 1)
    return a;
  else {
    int result = lgpow(a, b / 2);
    if(b % 2 == 0)
      return 1LL * result * result % modulo;
    else
      return 1LL * result * result % modulo * a % modulo;
  }
}

int mex(std::vector<int> v) {
  std::sort(v.begin(), v.end());
  v.erase(std::unique(v.begin(), v.end()), v.end());
  for(int i = 0; i < v.size(); i++)
    if(i != v[i])
      return i;
  return v.size();
}

int sum[3][5 + nmax];

int main() {
  int n, m[3];
  std::cin >> n;
  for(int h = 0; h < 3; h++) {
    std::cin >> m[h];
    for(int i = 1;i <= m[h]; i++){
      int x, y;
      std::cin >> x >> y;
      if(y < x)
        std::swap(x, y);
      g[h][x].push_back(y);
    }
    std::vector<int> cand;
    for(int i = n; 1 <= i; i--) {
      cand.clear();
      for(int h_ = 0; h_ < g[h][i].size(); h_++){
        int to = g[h][i][h_];
        cand.push_back(nimber[h][to]);
      }
      nimber[h][i] = mex(cand);
      sum[h][nimber[h][i]] = (sum[h][nimber[h][i]] + lgpow(10, 18 * i)) % modulo;
    }
  }

  int result = 0;
  
  for(int i = 0; i < rad; i++)
    for(int j = 0; j < rad; j++)
      if((i ^ j) < rad) { 
        result = (result + 1LL * sum[0][i] * sum[1][j] % modulo * sum[2][i ^ j]) % modulo;
      }
  std::cout << result;
}