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:
// by newbiechd
#include <bits/stdc++.h>
using namespace std;

#define debug(x) cerr << #x << " = " << x << endl

const int N_MAX = 100003;
int n;

const int mod = 998244353;
int powK[N_MAX];

struct Graph {
  int m;
  vector<int> g[N_MAX];
  bool tub[N_MAX];
  int sg[N_MAX], mx, f[N_MAX];
  
  void build() {
    cin >> m;
    for (int i = 1; i <= m; ++i) {
      int x, y;
      cin >> x >> y;
      if (x > y)
        swap(x, y);
      g[x].push_back(y);
    }
    for (int x = n; x > 0; --x) {
      for (int y : g[x])
        tub[sg[y]] = 1;
      int i = 0;
      while (tub[i])
        ++i;
      sg[x] = i;
      mx = max(mx, i);
      f[i] = (powK[x] + f[i]) % mod;
      for (int y : g[x])
        tub[sg[y]] = 0;
    }
  }
}G[3];

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  
  cin >> n;
  const int K = (long long)1e18 % mod;
  powK[0] = 1;
  for (int i = 1; i <= n; ++i)
    powK[i] = 1ll * K * powK[i - 1] % mod;
  for (int i = 0; i < 3; ++i)
    G[i].build();
  int ans = 0;
  for (int i = 0; i <= G[0].mx; ++i)
    for (int j = 0; j <= G[1].mx; ++j)
      ans = (1ll * G[0].f[i] * G[1].f[j] % mod * G[2].f[i ^ j] + ans) % mod;
  cout << ans << endl;
  return 0;
}