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;

const int md = 998244353;

vector<int> solveOne(int n) {
	vector<vector<int>> G(n);
	
	int m;
	ignore = scanf("%d", &m);
	for (int i = 0, u, v; i < m; i++) {
		ignore = scanf("%d %d", &u, &v);
		u--; v--;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	
	vector<int> w;
	w.push_back(1000000000LL * 1000000000LL % md);
	for (int i = 1; i < n; i++) {
		w.push_back(w.back() * 1LL * w.front() % md);
	}
	
	vector<int> res, grp(n);
	for (int i = n - 1; i >= 0; i--) {
		set<int> bad;
		for (int u : G[i]) {
			if (u < i) continue;
			bad.insert(grp[u]);
		}
		
		int g = 0;
		while (bad.count(g) == 1) g++;
		
		grp[i] = g;
		
		if (res.size() < g + 1) {
			res.resize(g + 1);
		}
		
		res[g] = (res[g] + w[i]) % md;
	}
	
	return res;
}

int main() {
	int n;
	ignore = scanf("%d", &n);
	
	auto v1 = solveOne(n);
	auto v2 = solveOne(n);
	auto v3 = solveOne(n);
	
	size_t n1 = v1.size();
	size_t n2 = v2.size();
	size_t n3 = v3.size();
	
	size_t N = max({n1, n2, n3});
	
	vector<vector<bool>> taken12(N, vector<bool>(N, false));
	auto taken13 = taken12;
	auto taken23 = taken12;
	
	int ans = 0;
	for (size_t s = 0; s < n1 + n2 + n3; s++)
		for (size_t i1 = 0; i1 < n1 && i1 <= s; i1++)
			for (size_t i2 = 0; i2 < n2 && i1 + i2 <= s; i2++) {
				size_t i3 = s - i1 - i2;
				if (i3 >= n3) continue;
				
				if (taken12[i1][i2]) continue;
				if (taken13[i1][i3]) continue;
				if (taken23[i2][i3]) continue;
				
				taken12[i1][i2] = true;
				taken13[i1][i3] = true;
				taken23[i2][i3] = true;
				
				ans = (ans + v1[i1] * 1LL * v2[i2] % md * v3[i3]) % md;
			}
	
	printf("%d\n", ans);
	
	return 0;
}