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 <cstdio>
#include <set>

const int P=998244353;
inline int mul(const int &a,const int &b){return 1ll*a*b%P;}
inline int add(int a,const int &b){a+=b;return(a>=P)?a-P:a;}
inline int sub(int a,const int &b){a-=b;return (a<0)?a+P:a;}
int qsm(int a,int b){
	int ans=1;
	while(b){
		if(b&1)ans=mul(ans,a);
		a=mul(a,a);
		b>>=1;
	}
	return ans;
}
int Pow[100001],f[131072],lim,n,head[100001],nxt[100001],b[100001],k,m,v[100001],g[131072];
void push(int s,int t){
	nxt[++k]=head[s];
	head[s]=k;
	b[k]=t;
}
std::set<int>set;
void FWT(int *a){
	for(int i=1;i<lim;i<<=1)
		for(int j=0;j<lim;j+=(i<<1))
			for(int k=0;k<i;k++){
				const int Ny=a[i+j+k];
				a[i+j+k]=sub(a[j+k],Ny);
				a[j+k]=add(a[j+k],Ny);
			}
}
int main(){
	scanf("%d",&n);
	lim=1;
	while(lim<=n)lim<<=1;
	for(int i=0;i<lim;i++)g[i]=1;
	Pow[0]=1;
	for(int i=1;i<=n;i++)Pow[i]=mul(Pow[i-1],(1000000000000000000ll)%P);
//	for(int i=1;i<=n;i++)printf("%d ",Pow[i]);putchar('\n');
	for(int fd=0;fd<3;fd++){
		for(int j=1;j<=n;j++)head[j]=v[j]=0;
		k=0;
		scanf("%d",&m);
		for(int j=1,u,v;j<=m;j++){
			scanf("%d%d",&u,&v);
			if(u>v)push(v,u);
			else push(u,v);
		}
		for(int i=n;i;i--){
			set.clear();
			for(int j=head[i];j;j=nxt[j])
				set.insert(v[b[j]]);
			for(v[i]=0;set.count(v[i]);v[i]++);
		}
		for(int j=0;j<lim;j++)f[j]=0;
		for(int j=1;j<=n;j++)f[v[j]]=add(f[v[j]],Pow[j]); 
		FWT(f);
		for(int j=0;j<lim;j++)g[j]=mul(g[j],f[j]);
	}
	int inv=qsm(lim,P-2),ans=0;
	for(int i=0;i<lim;i++)ans=add(ans,g[i]);
	printf("%d\n",mul(ans,inv));
}