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>
#include <array>
using namespace std;
using ULL = unsigned long long;
using UL = unsigned;
using LL = long long;
#define rep(i, n) for(UL i = 0; i < (n); i++)

template<class Ty>
using passive_queue = priority_queue<Ty, vector<Ty>, greater<Ty>>;

struct Problem {

	void Solve() {
		const ULL M = 998244353;
		const ULL Z = 1000000000000000000 % 998244353;


		UL N; cin >> N;

		vector<UL> POWZ(N + 1); POWZ[0] = 1;
		rep(i, N) POWZ[i + 1] = POWZ[i] * Z % M;
		vector<ULL> A[3];

		const UL X = min(N, 512u);

		rep(t, 3) {
			UL S; cin >> S;
			vector<vector<UL>> E(N);
			rep(j, S) {
				UL u, v; cin >> u >> v; u--; v--;
				E[min(u, v)].push_back(max(u, v));
			}
			A[t].resize(X);
			vector<UL> F(X, -1), G(N);
			for (UL v = N - 1; v != -1; v--) {
				for (UL e : E[v]) F[G[e]] = v;
				while (F[G[v]] == v) G[v]++;
				A[t][G[v]] += POWZ[v + 1];
				A[t][G[v]] %= M;
			}
		}

		ULL ans = 0;
		{
			rep(a, X) rep(b, X) {
				UL c = a ^ b; if (c >= X) continue;
				ans += A[0][a] * A[1][b] % M * A[2][c] % M;
				ans %= M;
			}
		}

		cout << ans << endl;
	}

	Problem();
};
int main() {
	unique_ptr<Problem> p(new Problem());
	p->Solve();
	return 0;
}
Problem::Problem() {
	cout << fixed << setprecision(10);
}