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;
 
const int S = 500, SSS = S * S * S, N = 1e5 + 5, mod = 998244353;
 
long long powmod(long long b, long long p) {
  long long r = 1;
  for (; p; p >>= 1, b = b * b % mod)
    if (p & 1)
      r = r * b % mod;
  return r;
}
 
const int base = powmod(10, 18);
 
vector<int> g[N];
 
vector<pair<int, int>> independent(int n, vector<pair<int ,int>> & edges) {
  for (int i = 0; i < n; ++i)
    g[i].clear();
  for (auto e : edges) {
    g[e.first].push_back(e.second);
    g[e.second].push_back(e.first);
  }
  vector<bool> done(n, false), mark(n);
  int cnt = 0;
  vector<pair<int, int>> res;
  while (cnt < n) {
    mark = done;
    int hig = -1, sum = 0;
    for (int v = n-1; v >= 0; --v) {
      if (mark[v]) continue;
      ++cnt;
      done[v] = 1;
      hig = max(hig, v+1);
      sum = (sum + powmod(base, v+1)) % mod;
      for (int u : g[v]) {
        mark[u] = 1;
      }
    }
    res.emplace_back(hig, sum);
  }
  return res;
}
 
vector<pair<int, int>> read_edges() {
  int m;
  scanf("%d", &m);
  vector<pair<int, int>> edges(m);
  for (int i = 0; i < m; ++i) {
    scanf("%d %d", &edges[i].first, &edges[i].second);
    --edges[i].first;
    --edges[i].second;
  }
  return edges;
}
 
int main() {
  int n;
  scanf("%d", &n);
  vector<pair<int, int>> ex = read_edges();
  vector<pair<int, int>> ey = read_edges();
  vector<pair<int, int>> ez = read_edges();
  vector<pair<int, int>> rx = independent(n, ex);
  vector<pair<int, int>> ry = independent(n, ey);
  vector<pair<int, int>> rz = independent(n, ez);
  assert(rx.size() < S);
  assert(ry.size() < S);
  assert(rz.size() < S);
  long long ans = 0;
  for (int i = 0; i < (int)rx.size(); ++i) {
    for (int j = 0; j < (int)ry.size(); ++j) {
      int k = i^j;
      if (k < (int)rz.size()) {
        long long cur = rx[i].second * (long long)ry[j].second % mod;
        ans = (ans + cur * rz[k].second) % mod;
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}