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
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=1e5+5,mod=998244353;
const int W=(long long)1e18%mod;
int ksm(int b,int n){
	int res=1;
	while(n){
		if(n&1) res=res*b%mod;
		b=b*b%mod; n>>=1;
	}
	return res;
}
int n;
struct graph{
	vector<int> e[N];
	int sg[N],val[N],mx;
	int SG(int u){
		if(sg[u]!=-1) return sg[u];
		map<int,bool> fl;
		for(int i=0;i<e[u].size();++i)
			fl[SG(e[u][i])]=1;
     	do{
          ++sg[u];
        }while(fl.count(sg[u]));
		return sg[u];
	}
	void solve(){
		int m,u,v;
		cin>>m;
		for(int i=1;i<=m;++i){
			cin>>u>>v;
			if(u>v) swap(u,v);
			e[u].push_back(v);
		}
		for(int i=1;i<=n;++i) sg[i]=-1;
		for(int i=1;i<=n;++i)
			if(sg[i]==-1) SG(i);
		for(int i=1;i<=n;++i)
			mx=max(mx,sg[i]),
			val[sg[i]]+=ksm(W,i),val[sg[i]]%=mod;
	//	for(int i=1;i<=n;++i)
		//	cout<<i<<" "<<sg[i]<<endl;
	}
}g0,g1,g2;
signed main(){
//	cout<<W<<endl;
	cin>>n;
	g0.solve();g1.solve();g2.solve();
	int ans=0;
	for(int i=0;i<=g0.mx;++i)
		for(int j=0;j<=g1.mx;++j)
			ans+=g0.val[i]*g1.val[j]%mod*g2.val[i^j]%mod,ans%=mod;
	cout<<ans;
	return 0;
}