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;

typedef long long ll;
typedef unordered_map <int, int> umap;

const int MAXN = 1e5 + 5;
const int MOD = 998244353;

int N;
umap G1, G2, G3;
vector <int> adj[MAXN];
int pot[MAXN];
int mex[MAXN];

inline void add(int &x, int y) {
	x += y;
	if (x >= MOD)
		x -= MOD;
}

inline int mul(int x, int y) {
	return (ll)x * y % MOD;
}

int dfs(int x) {
	if (mex[x] != -1)
		return mex[x];
	umap occ;
	for (auto it : adj[x])
		occ[dfs(it)] = 1;
	occ[-1] = 1;
	while (occ[mex[x]])
		mex[x]++;
	return mex[x];
}

void input(umap &G) {
	memset(mex, -1, sizeof mex);
	for (int i = 1; i <= N; i++)
		adj[i].clear();
	int M;
	scanf("%d", &M);
	while (M--) {
		int a, b;
		scanf("%d%d", &a, &b);
		adj[min(a, b)].push_back(max(a, b));
	}
	for (int i = 1; i <= N; i++)
		add(G[dfs(i)], pot[i]);
}

void load() {
	scanf("%d", &N);
	pot[0] = 1;
	for (int i = 1; i <= N; i++)
		pot[i] = mul(pot[i - 1], (ll)1e18 % MOD);
	input(G1);
	input(G2);
	input(G3);
}

int solve() {
	umap conv;
	for (auto it1 : G1)
		for (auto it2 : G2)
			add(conv[it1.first ^ it2.first], mul(it1.second, it2.second));
	int sol = 0;
	for (auto it : G3)
		add(sol, mul(conv[it.first], it.second));
	return sol;
}

int main() {
	load();
	printf("%d\n", solve());
	return 0;
}