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 int LL;

#define st first
#define nd second
#define PLL pair <LL, LL>
#define PII pair <int, int>

const int S = 2048;
const int N = 5e5 + 7;
const int MX = 998244353;

int n;
vector <int> G[N];

int last[N];
int color[N];

int dp[N];
int cur[N];

int tmp[N];
int pre[N];

int getMagic(){
	int cur = 1000 * 1000 * 1000 % MX;
	return 1LL * cur * cur % MX;
}

void add(int &a, int b){
	a += b;
	if(a >= MX)
		a -= MX;
}

void getMex(int u){
	for(auto v: G[u])
		if(v > u)
			last[color[v]] = u;
	
	for(int t = 0; true; ++t)
		if(last[t] != u){
			color[u] = t;
			break;
		}
}

void solve(int id){
	int m;
	scanf("%d", &m);
	
	last[0] = n + 1;
	for(int i = 1; i <= n; ++i){
		G[i].clear();
		color[i] = 0;
		last[i] = n + 1;
	}
	
	for(int i = 1; i <= m; ++i){
		int u, v;
		scanf("%d %d", &u, &v);
		
		G[u].push_back(v);
		G[v].push_back(u);
	}
	
	for(int i = 0; i < S; ++i)
		cur[i] = 0;
	
	for(int i = n; i >= 1; --i){
		getMex(i);
		add(cur[color[i]], pre[i]);
	}
	
	for(int i = 0; i < S; ++i){
		tmp[i] = dp[i];
		dp[i] = 0;
	}
	
	for(int i = 0; i < S; ++i)
		for(int j = 0; j < S; ++j)
			add(dp[i ^ j], 1LL * tmp[i] * cur[j] % MX);
}

int main(){
	pre[0] = 1;
	for(int i = 1; i < N; ++i)
		pre[i] = 1LL * pre[i - 1] * getMagic() % MX;
	dp[0] = 1;

	scanf("%d", &n);
	for(int t = 1; t <= 3; ++t)
		solve(t);
	
	printf("%d\n", dp[0]);
	return 0;
}