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:
// -----------------------------------
// Author     : MatsuTaku
// Affiliation: Tokushima University
// Country    : Japan
// Date       : 03/21/2020
// -----------------------------------

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

constexpr ll MOD = 998244353;
constexpr ll MAX_GRANDY = 512;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);

  int N; cin>>N;
  vector<int> graph[3][N+1];
  for (int i = 0; i < 3; i++) {
    int M; cin>>M;
    for (int m = 0; m < M; m++) {
      int a,b; cin>>a>>b;
      if (a > b) swap(a,b);
      graph[i][a].push_back(b);
    }
  }

  vector<ll> pw(N+1);
  pw[0] = 1;
  constexpr ll p = ll(1e18)%MOD;
  for (int i = 1; i <= N; i++) {
    pw[i] = pw[i-1]*p%MOD;
  }

  array<array<ll, MAX_GRANDY>, 3> cnt={};
  for (int i = 0; i < 3; i++) {
    vector<int> gr(N+1, 0);
    for (int v = N; v >= 1; v--) {
      bitset<MAX_GRANDY> ps;
      for (auto t : graph[i][v])
        ps[gr[t]] = true;
      while (ps[gr[v]])
        ++gr[v];
      cnt[i][gr[v]] += pw[v];
      cnt[i][gr[v]] %= MOD;
    }
  }

  ll ans = 0;
  for (int i = 0; i < MAX_GRANDY; i++) for (int j = 0; j < MAX_GRANDY; j++) {
    ans += cnt[0][i] * cnt[1][j] % MOD * cnt[2][i^j] % MOD;
    ans %= MOD;
  }
  cout << ans << endl;

  return 0;
}