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"
using namespace std;

constexpr long long MOD = 998244353;
constexpr long long U = 1000000000000000000 % MOD;

void input(vector<vector<int> >& V) {
	int M;
	cin >> M;
	for (int i = 0; i < M; i++) {
		int A, B;
		cin >> A >> B;
		A--, B--;
		if (A > B) swap(A, B);
		V[A].push_back(B);
	}
}

void Grundy(vector<int>& G, vector<vector<int> >& V) {
	for (int i = (int)V.size() - 1; i >= 0; i--) {
		if (V[i].empty()) {
			G[i] = 0;
			continue;
		}
		vector<int> Gs;
		for (int j : V[i]) {
			Gs.push_back(G[j]);
		}
		sort(Gs.begin(), Gs.end());
		Gs.erase(unique(Gs.begin(), Gs.end()), Gs.end());
		if (Gs[Gs.size() - 1] == Gs.size() - 1) {
			G[i] = Gs.size();
			continue;
		}
		for (int j = 0; j < Gs.size(); j++) {
			if (Gs[j] != j) {
				G[i] = j;
				break;
			}
		}
	}
}

long long Pow(int A) {
	if (A == 0) return 1;
	if (A % 2 == 0) {
		long long B = Pow(A / 2);
		return (B * B) % MOD;
	}
	return (U * Pow(A - 1)) % MOD;
}

void Count(vector<long long>& C, vector<int>& G) {
	for (int i = 0; i < G.size(); i++) {
		C[G[i]] += Pow(i + 1);
		C[G[i]] %= MOD;
	}
}

int main() {
	int N;
	cin >> N;
	vector<vector<int> > X(N), Y(N), Z(N);
	vector<int> GX(N), GY(N), GZ(N);
	vector<long long> CX(512, 0), CY(512, 0), CZ(512, 0);
	input(X), input(Y), input(Z);
	Grundy(GX, X), Grundy(GY, Y), Grundy(GZ, Z);
	Count(CX, GX), Count(CY, GY), Count(CZ, GZ);
	long long ANS = 0;
	for (int i = 0; i < 512; i++) {
		for (int j = 0; j < 512; j++) {
			long long MUL = (CX[i] * CY[j]) % MOD;
			MUL *= CZ[i ^ j];
			MUL %= MOD;
			ANS += MUL;
			ANS %= MOD;
		}
	}
	cout << ANS << endl;
}