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;

typedef long long ll;

const int N = 500010;
const int MOD = 998244353;
const ll INF = 1000000LL * 1000000LL * 1000000LL % MOD;

vector <int> g[N];
ll pwr[N], sum[3][N];
int n, m, grundy[N], lim;

int main() {
  pwr[0] = 1;
  for (int i = 1; i < N; ++i) pwr[i] = INF * pwr[i - 1] % MOD;
  cin >> n;
  for (int it = 0; it < 3; ++it) {
    cin >> m;
    while (m--) {
      int u, v;
      scanf("%d %d", &u, &v);
      g[u].emplace_back(v);
      g[v].emplace_back(u);
    }
    for (int i = n; i >= 1; --i) {
      set <int> nei;
      for (int j : g[i]) if (j > i) nei.emplace(grundy[j]);
      grundy[i] = 0;
      for (int x : nei) {
        if (x == grundy[i]) ++grundy[i]; 
        else break;
      }
      lim = max(lim, grundy[i]);
      sum[it][grundy[i]] += pwr[i], sum[it][grundy[i]] %= MOD;
    }
    for (int i = 1; i <= n; ++i) g[i].clear();
  }
  ll ans = 0;
  for (int i = 0; i <= lim; ++i) {
    for (int j = 0; j <= lim; ++j) {
      ll cur = (sum[0][i] * sum[1][j] % MOD) * sum[2][i ^ j] % MOD;
      ans += cur, ans %= MOD;
    }
  }
  cout << ans << '\n';
  return 0;
}

