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;
template<typename T>inline T read(){
	T f=0,x=0;char c=getchar();
	while(!isdigit(c)) f=c=='-',c=getchar();
	while(isdigit(c)) x=x*10+c-48,c=getchar();
	return f?-x:x;
}
namespace run{
	typedef pair<int,int> P;
	const int N=1e5+9,mod=998244353,inv2=(mod+1)/2;
	int head[N],nex[N],to[N],cnt;
	inline void add(int u,int v){
		nex[++cnt]=head[u];
		head[u]=cnt;to[cnt]=v;
	}

	int n,m,pw[N],sg[N],f[1<<20],limit=1,ans[1<<20];
	inline void fwt_xor(int *a,int opt){
		for(int mid=1;mid<limit;mid<<=1)
			for(int i=0,R=mid<<1;i<limit;i+=R)
				for(int j=0;j<mid;j++){
					int x=a[i+j],y=a[i+j+mid];
					a[i+j]=1LL*(x+y)*(opt?1:inv2)%mod;
					a[i+j+mid]=(1LL*(x-y)*(opt?1:inv2)%mod+mod)%mod;
				}
	}
	map<P,bool> mp;
	int main(){
		n=read<int>();
		while(limit<=n) limit<<=1;
		int bas=((long long)1e18)%mod;pw[0]=1;
		for(int i=1;i<=n;i++) pw[i]=1LL*pw[i-1]*bas%mod;
		for(int i=0;i<limit;i++) ans[i]=1;
		for(int num=0;num<3;num++){
			m=read<int>();
			for(int i=1;i<=n;i++) head[i]=0,sg[i]=0;cnt=0;
			for(int i=1;i<=m;i++){
				int u=read<int>(),v=read<int>();
				add(min(u,v),max(u,v));
			}
			mp.clear();
			for(int u=n;u>=1;u--){
				for(int i=head[u];i;i=nex[i])
					if(~sg[to[i]]) mp[P(u,sg[to[i]])]=1;
				while(mp[P(u,sg[u])]) sg[u]++;
			}
		//	for(int i=1;i<=n;i++)
		//		cout<<sg[i]<<endl;
			for(int i=0;i<limit;i++) f[i]=0;
			for(int i=1;i<=n;i++) f[sg[i]]=(f[sg[i]]+pw[i])%mod;
			fwt_xor(f,1);
			for(int i=0;i<limit;i++) ans[i]=1LL*ans[i]*f[i]%mod;
		}
		fwt_xor(ans,0);
		printf("%d\n",ans[0]);
		return 0;
	}
}
int main(){
#ifdef my
	freopen("graph.in","r",stdin);
	freopen("graph.out","w",stdout);
#endif
	return run::main();
}