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 <bits/stdc++.h>
using namespace std;

const int MOD = 998244353;

int Power(int n, int x) {
  if (x == 0) return 1;
  int res = Power(n, x / 2);
  res = 1ll * res * res % MOD;
  if (x & 1) res = 1ll * res * n % MOD;
  return res;
}

const int INV = Power(Power(10, 18), MOD - 2);

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);

  int N;
  cin >> N;
  vector<vector<int>> contrib(3);
  for (int g = 0; g < 3; g++) {
    int M;
    cin >> M;
    vector<vector<int>> adj(N);
    for (int i = 0; i < M; i++) {
      int u, v;
      cin >> u >> v;
      u--, v--;
      if (u > v) {
        swap(u, v);
      }
      adj[u].emplace_back(v);
    }
    int cur = Power(10, 18 * N);
    vector<int> used(N);
    vector<int> grundy(N);
    for (int i = N - 1; i >= 0; i--) {
      for (auto j : adj[i]) {
        used[grundy[j]] = 1;
      }
      while (used[grundy[i]]) grundy[i]++;
      while (int(contrib[g].size()) <= grundy[i]) {
        contrib[g].push_back(0);
      }
      contrib[g][grundy[i]] += cur;
      contrib[g][grundy[i]] %= MOD;
      cur = 1ll * cur * INV % MOD;
      for (auto j : adj[i]) {
        used[grundy[j]] = 0;
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < (int) contrib[0].size(); i++) {
    for (int j = 0; j < (int) contrib[1].size(); j++) {
      int k = i ^ j;
      if (k < (int) contrib[2].size()) {
        ans += 1ll * contrib[0][i] * contrib[1][j] % MOD * contrib[2][k] % MOD;
        ans %= MOD;
      }
    }
  }
  cout << ans << "\n";
  return 0;
}