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
#include <bits/stdc++.h>

using namespace std;

using ll = long long;
const int MOD = 998244353;
const int BASE = 716070898;

int main() {
   ios_base::sync_with_stdio(false);
   int N;
   cin >> N;
   vector<int> pw(N + 1);
   pw[0] = 1;
   for (int i = 0; i < N; ++i) {
      pw[i + 1] = ll(pw[i]) * BASE % MOD;
   }
   vector<array<int, 512>> cnts;
   for (int z = 0; z < 3; ++z) {
      int M;
      cin >> M;
      vector<vector<int>> adj(N);
      for (int i = 0; i < M; ++i) {
         int v, u;
         cin >> v >> u;
         --v, --u;
         if (v > u) swap(v, u);
         adj[v].emplace_back(u);
      }
      array<int, 512> cnt = {};
      vector<int> sg(N);
      for (int v = N - 1; v >= 0; --v) {
         set<int> nxts;
         for (int u : adj[v]) {
            nxts.emplace(sg[u]);
         }
         sg[v] = 0;
         while (nxts.count(sg[v])) sg[v]++;
         cnt[sg[v]] += pw[v + 1];
         if (cnt[sg[v]] >= MOD) cnt[sg[v]] -= MOD;
      }
      cnts.emplace_back(cnt);
   }
   int ans = 0;
   for (int x = 0; x < 512; ++x) {
      for (int y = 0; y < 512; ++y) {
         ans += ll(cnts[0][x]) * cnts[1][y] % MOD * cnts[2][x ^ y] % MOD;
         if (ans >= MOD) ans -= MOD;
      }
   }
   cout << ans << "\n";
   return 0;
}
