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<algorithm>
#include<iostream>
#include<climits>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#define ll long long
using namespace std;
//{{{ read()
inline int read(){
	register int x=0,f=1;
	register char ch=getchar();
	while(ch<'0'||ch>'9'){
		if(ch=='-')	f=-1;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9')	x=x*10+(ch^48),ch=getchar();
	return x*f;
}
//}}}
const int N=1e5+5;
const int P=998244353;
const int d=1000000000000000000ll%P;
vector<int>A[N],B[N],C[N];
int n,m,mx,f[N],a[N],b[N],f1[N],f2[N],f3[N],vi[N];
void Add(int &x,int y){
	x+=y,x-=x>=P?P:0;
}
ll fsp(ll x,int y){
	ll ans=1;
	while(y){
		if(y&1)	ans=ans*x%P;
		x=x*x%P,y>>=1;
	}
	return ans;
}
int main(){
	n=read();
	mx=max(mx,m=read());	while(m--){
		int x=read(),y=read();
		if(x>y)	swap(x,y);	
		A[x].push_back(y);
	}
	mx=max(mx,m=read());	while(m--){
		int x=read(),y=read();
		if(x>y)	swap(x,y);	
		B[x].push_back(y);
	}
	mx=max(mx,m=read());	while(m--){
		int x=read(),y=read();
		if(x>y)	swap(x,y);	
		C[x].push_back(y);
	}
	for(int i=n;i;i--){
		int k=0,si=A[i].size();
		for(int j=0;j<si;j++)	vi[f1[A[i][j]]]=1;
		while(vi[k])	++k;
		f1[i]=k;
		for(int j=0;j<si;j++)	vi[f1[A[i][j]]]=0;
	}
	for(int i=n;i;i--){
		int k=0,si=B[i].size();
		for(int j=0;j<si;j++)	vi[f2[B[i][j]]]=1;
		while(vi[k])	++k;
		f2[i]=k;
		for(int j=0;j<si;j++)	vi[f2[B[i][j]]]=0;
	}
	for(int i=n;i;i--){
		int k=0,si=C[i].size();
		for(int j=0;j<si;j++)	vi[f3[C[i][j]]]=1;
		while(vi[k])	++k;
		f3[i]=k;
		for(int j=0;j<si;j++)	vi[f3[C[i][j]]]=0;
	}
	for(int i=1;i<=n;i++)	Add(a[f1[i]],fsp(d,i));
	for(int i=1;i<=n;i++)	Add(b[f2[i]],fsp(d,i));
	int k=sqrt(mx)*2;
	for(int i=0;i<=k;i++)	
		for(int j=0;j<=k;j++)
			Add(f[i^j],1LL*a[i]*b[j]%P);
	for(int i=0;i<=k;i++)	a[i]=f[i],f[i]=0,b[i]=0;
	for(int i=1;i<=n;i++)	Add(b[f3[i]],fsp(d,i));
	for(int i=0;i<=k;i++)	
		for(int j=0;j<=k;j++)
			Add(f[i^j],1LL*a[i]*b[j]%P);
	printf("%d\n",f[0]);
	return 0;
}
