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;

#define sz(x) ((int) ((x).size()))
typedef long long ll;
typedef long double ld;

const ll mod = 998244353, p18 = (ll) 1e18 % mod;
const int mg = 1 << 9;
int n, m, g[100100];
ll p[100100], r[mg];
vector<int> e[100100];

void solve() {
  cin >> m;
  for (int i = 0; i < n; i++)
    e[i].clear();
  int a, b;
  for (int i = 0; i < m; i++) {
    cin >> a >> b, a--, b--;
    if (a > b)
      swap(a, b);
    e[a].push_back(b);
  }
  ll s[mg], nr[mg];
  for (int i = 0; i < mg; i++)
    s[i] = nr[i] = 0;
  for (int i = n - 1; i >= 0; i--) {
    sort(e[i].begin(), e[i].end(), [&](int a, int b){return g[a] < g[b];});
    reverse(e[i].begin(), e[i].end());
    g[i] = 0;
    while (!e[i].empty() && g[e[i].back()] == g[i]) {
      g[i]++;
      while (!e[i].empty() && g[e[i].back()] < g[i])
        e[i].pop_back();
    }
    s[g[i]] += p[i + 1];
  }
  for (int i = 0; i < mg; i++)
    s[i] %= mod;
  for (int i = 0; i < mg; i++)
    for (int j = 0; j < mg; j++)
      nr[i ^ j] = (nr[i ^ j] + s[i] * r[j]) % mod;
  for (int i = 0; i < mg; i++)
    r[i] = nr[i];
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(10);
  cin >> n;
  p[0] = 1;
  for (int i = 1; i <= n; i++)
    p[i] = (p[i - 1] * p18) % mod;
  r[0] = 1;
  solve();
  solve();
  solve();
  cout << r[0] << "\n";
}