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>
#define SIZE 100005
#define SQR 512
#define MOD 998244353
#define INF 1000000000000000000LL
using namespace std;
typedef long long int ll;

int n;
vector <int> vec[SIZE];
int grundy[SIZE];
ll sum[3][SQR];

void solve(int t)
{
	int m;
	scanf("%d",&m);
	for(int i=0;i<n;i++) vec[i].clear();
	for(int i=0;i<m;i++)
	{
		int a,b;
		scanf("%d %d",&a,&b);
		a--,b--;
		if(a>b) swap(a,b);
		vec[a].push_back(b);
	}
	for(int i=n-1;i>=0;i--)
	{
		set <int> st;
		for(int j=0;j<vec[i].size();j++) st.insert(grundy[vec[i][j]]);
		grundy[i]=0;
		while(st.count(grundy[i])) grundy[i]++;
	}
	ll mul=1;
	for(int i=0;i<n;i++)
	{
		mul*=INF%MOD;
		mul%=MOD;
		sum[t][grundy[i]]+=mul;
		if(sum[t][grundy[i]]>=MOD) sum[t][grundy[i]]-=MOD;
	}
}
int main()
{
	scanf("%d",&n);
	for(int t=0;t<3;t++) solve(t);
	ll ret=0;
	for(int i=0;i<SQR;i++)
	{
		for(int j=0;j<SQR;j++)
		{
			ret+=sum[0][i]*sum[1][j]%MOD*sum[2][i^j]%MOD;
			if(ret>=MOD) ret-=MOD;
		}
	}
	printf("%lld\n",ret);
	return 0;
}