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 pair<ll,ll> P;

#define fr first
#define sc second
#define chmax(a,b) a=max(a,b)
#define chmin(a,b) a=min(a,b)

const ll MOD=998244353;

ll x[100010];
void init(){
	x[0]=1e+18;
	//x[0]=2;
	x[0]%=MOD;
	for(int i=1;i<100010;i++){
		x[i]=x[i-1]*x[0];
		x[i]%=MOD;
	}
}

ll modpow(ll x,ll k){
	if(k==0)return 1;
	ll ret=modpow(x,k/2);
	ret*=ret;
	ret%=MOD;
	if(k%2==1){
		ret*=x;
		ret%=MOD;
	}
	return ret;
}

struct Graph{
	int n,m;
	vector<int> G[100010];
	int k;
	ll ret[100010];
	bool used[100010];
	bool used2[100010];
	int run(){
		memset(used,0,sizeof used);
		int zan=n;
		k=0;
		while(zan>0){
			ret[k]=0;
			memset(used2,0,sizeof used2);
			for(int i=n;i>=1;i--){
				if(used[i])continue;
				used2[i]=true;
				for(int v: G[i]){
					if(used2[v]){
						used2[i]=false;
						break;
					}
				}
				if(used2[i]){
					ret[k]+=x[i-1];
					if(ret[k]>=MOD)ret[k]-=MOD;
					zan--;
					used[i]=true;
					//cerr<<i<<" "<<k<<" "<<x[i]<<" "<<ret[k]<<endl;
				}
			}
			k++;
		}
		/*cerr<<k<<endl;
		for(int i=0;i<k;i++){
			cerr<<ret[i]<<" ";
		}cerr<<endl;*/
		return k;
	}
}G[3];

int main(){
	init();
	scanf("%d",&G[0].n);
	G[1].n=G[0].n;
	G[2].n=G[0].n;
	int k=MOD;
	for(int t=0;t<3;t++){
		scanf("%d",&G[t].m);
		for(int i=0;i<G[t].m;i++){
			int a,b;
			scanf("%d%d",&a,&b);
			if(a>b)G[t].G[b].push_back(a);
			else G[t].G[a].push_back(b);
		}
		chmin(k,G[t].run());
	}
	
	static ll a[1<<17];
	for(int i=0;i<G[1].k;i++){
		for(int j=0;j<G[2].k;j++){
			a[i^j]+=G[1].ret[i]*G[2].ret[j];
			a[i^j]%=MOD;
		}
	}
	
	ll ret=0;
	for(int i=0;i<G[0].k;i++){
		a[i]*=G[0].ret[i];
		ret+=a[i]; ret%=MOD;
	}
	cout<<ret%MOD<<endl;
}