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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define MOD 998244353
#define ADD(X,Y) ((X) = ((X) + (Y)%MOD) % MOD)
typedef long long i64; typedef vector<int> ivec; typedef vector<string> svec;

i64 mof = 1000000000000000000LL % MOD;

int N;
vector<int> edge[101010];
int grundy[101010];

vector<i64> analyze()
{
	int M;
	scanf("%d", &M);
	for (int i = 1; i <= N; ++i) edge[i].clear();
	for (int i = 0; i < M; ++i) {
		int a, b;
		scanf("%d%d", &a, &b);
		if (a > b) swap(a, b);
		edge[a].push_back(b);
	}
	grundy[N] = 0;
	for (int i = N - 1; i >= 1; --i) {
		set<int> waf;
		for (int j : edge[i]) waf.insert(grundy[j]);
		for (int j = 0;; ++j) {
			if (waf.count(j) == 0) {
				grundy[i] = j;
				break;
			}
		}
	}
	vector<i64> res(1024, 0LL);

	i64 w = 1;
	for (int i = 1; i <= N; ++i) {
		w = w * mof % MOD;
		ADD(res[grundy[i]], w);
	}
	return res;
}

int main()
{
	scanf("%d", &N);
	vector<i64> seq1 = analyze();
	vector<i64> seq2 = analyze();
	vector<i64> seq3 = analyze();
	i64 ret = 0;
	for (int i = 0; i < 1024; ++i) {
		for (int j = 0; j < 1024; ++j) {
			ADD(ret, seq1[i] * seq2[j] % MOD * seq3[i ^ j]);
		}
	}
	printf("%lld\n", ret);
	return 0;
}