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;
#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define RFOR(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define REP(i,n) for (int i=0;i<(n);i++)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)
typedef long long LL;
int N;
int M[3];
vector<int>G[3][100001];
bool used[3][100001];
LL gra[3][100001];
LL ans[3][100001];

LL MOD=998244353;
LL MM(LL x){
	LL answer=1;
	LL now=10;
	while(x!=0){
		if(x%2==1){
			answer*=now;
			answer%=MOD;
		}
		now*=now;
		now%=MOD;
		x/=2;
	}
	return answer;
}

void calc(int x){
	for(int i=N-1;i>=0;i--){
		int now=0;
		vector<int>v;
		REP(j,G[x][i].size()){
			if(G[x][i][j]>i){
				v.push_back(gra[x][G[x][i][j]]);
			}
		}
		sort(v.begin(),v.end());
		REP(j,v.size()){
			if(v[j]==now)now++;
		}
		gra[x][i]=now;
		ans[x][now]+=MM(18*(i+1));
		ans[x][now]%=MOD;
	}
}



int main(){
	cin>>N;
	REP(i,3){
		cin>>M[i];
		REP(j,M[i]){
			int a,b;
			cin>>a>>b;
			a--;
			b--;
			G[i][a].push_back(b);
			G[i][b].push_back(a);
		}
	}
	REP(i,3){
		REP(j,N){
			used[i][j]=false;
		}
	}
	REP(i,3){
		calc(i);
	}
	/*REP(i,3){
		REP(j,N){
			cout<<ans[i][j]<<endl;
		}
		cout<<endl;
	}*/
	LL answer=0;
	REP(i,1001){
		REP(j,1001){
			int now=i^j;
			LL temp=ans[0][i];
			temp%=MOD;
			temp*=ans[1][j];
			temp%=MOD;
			temp*=ans[2][now];
			temp%=MOD;
			//LL temp=18*(ans[0][i]*ans[1][j]*ans[2][now]);
			/*if(temp!=0){
				cout<<i<<" "<<j<<" "<<now<<" "<<temp<<endl;
				cout<<ans[0][i]<<"ans "<<ans[1][j]<<" "<<ans[2][now]<<" "<<temp<<endl;
			}*/
			answer+=temp;
			answer%=MOD;
		}
	}
	cout<<answer<<endl;
}