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<algorithm>
#include<cctype>
#include<cstring>
using namespace std;
#define G getchar()
int read()
{
	int x=0; char ch=G;
	for (;!isdigit(ch);ch=G);
	for (;isdigit(ch);ch=G) x=(x<<3)+(x<<1)+(ch^48);
	return x;
}
#undef G
typedef long long ll;
const ll INF=1e18;
const int Mod=998244353;

int n,m; int p[100010];
struct Edge{
	int to,nxt;
}edge[100010];
int cnt,last[100010];
inline void addedge(int x,int y){
	if (x>y) swap(x,y);
	edge[++cnt]=(Edge){y,last[x]},last[x]=cnt;
}
int mxa=-1,mxb=-1,mxc=-1;
int a[100010],b[100010],c[100010];
bool mrk[100010]; int sg[100010];
int main()
{
	n=read();
	p[0]=1; p[1]=INF%Mod; for (int i=2;i<=n;i++) p[i]=1LL*p[i-1]*p[1]%Mod;
	
	m=read();
	cnt=0; memset(last+1,0,n*sizeof(int));
	for (int i=1;i<=m;i++) addedge(read(),read());
	for (int i=n;i;i--){
		for (int j=last[i],v;j;j=edge[j].nxt) mrk[sg[v=edge[j].to]]=1;
		for (sg[i]=0;mrk[sg[i]];sg[i]++);
		a[sg[i]]=(a[sg[i]]+p[i])%Mod; mxa=max(mxa,sg[i]);
		for (int j=last[i],v;j;j=edge[j].nxt) mrk[sg[v=edge[j].to]]=0;
	}
	
	m=read();
	cnt=0; memset(last+1,0,n*sizeof(int));
	for (int i=1;i<=m;i++) addedge(read(),read());
	for (int i=n;i;i--){
		for (int j=last[i],v;j;j=edge[j].nxt) mrk[sg[v=edge[j].to]]=1;
		for (sg[i]=0;mrk[sg[i]];sg[i]++);
		b[sg[i]]=(b[sg[i]]+p[i])%Mod; mxb=max(mxb,sg[i]);
		for (int j=last[i],v;j;j=edge[j].nxt) mrk[sg[v=edge[j].to]]=0;
	}
	
	m=read();
	cnt=0; memset(last+1,0,n*sizeof(int));
	for (int i=1;i<=m;i++) addedge(read(),read());
	for (int i=n;i;i--){
		for (int j=last[i],v;j;j=edge[j].nxt) mrk[sg[v=edge[j].to]]=1;
		for (sg[i]=0;mrk[sg[i]];sg[i]++);
		c[sg[i]]=(c[sg[i]]+p[i])%Mod; mxc=max(mxc,sg[i]);
		for (int j=last[i],v;j;j=edge[j].nxt) mrk[sg[v=edge[j].to]]=0;
	}
	
	int ans=0;
	for (int i=0;i<=mxa;i++)
		for (int j=0;j<=mxb;j++)
			if ((i^j)<=mxc)
				ans=(ans+1LL*a[i]*b[j]%Mod*c[i^j])%Mod;
	printf("%d\n",(ans+Mod)%Mod);
	return 0;
}