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<cstdio>
#include<iostream>
#include<algorithm>

#define P 998244353
#define X 716070898

using namespace std;

int read(){
	int x=0;
	char ch=getchar();
	while(!isdigit(ch)) ch=getchar();
	while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	return x;
}

const int N = 100005;

int n,m[3];
struct node{
	int v;
	node *nxt;
}pool[N*3],*h[3][N];
int cnt;
void addedge(int ty,int u,int v){
	node *p=&pool[++cnt];
	p->v=v;p->nxt=h[ty][u];h[ty][u]=p;
}

int mx[3],sum[3][N],used[N],sg[N],mul[N];
int Plus(int x,int y) { return x+y>=P ? x+y-P : x+y; }
void cal(int ty){
	for(int i=0;i<=m[ty];i++) used[i]=0;
	int cur=0;
	sg[n]=0; sum[ty][0]=Plus(sum[ty][0],mul[n]);
	for(int i=n-1;i>0;i--){
		++cur;
		for(node *p=h[ty][i];p;p=p->nxt) used[sg[p->v]]=cur;
		sg[i]=-1;
		for(int j=0;j<=mx[ty];j++) 
			if(used[j]!=cur){ sg[i]=j; break; }
		if(sg[i]==-1) sg[i]=++mx[ty];
		sum[ty][sg[i]]=Plus(sum[ty][sg[i]],mul[i]);
	}
}

int main()
{
	int u,v;
	n=read();
	for(int i=0;i<3;i++){
		m[i]=read();
		for(int j=0;j<m[i];j++){
			u=read(); v=read();
			if(u>v) swap(u,v);
			addedge(i,u,v);
		}
	}
	
	mul[0]=1;
	for(int i=1;i<=n;i++) mul[i]=1ll*mul[i-1]*X%P;
	cal(0); cal(1); cal(2);
	
	int ans=0;
	for(int i=0;i<=mx[0];i++)
		for(int j=0;j<=mx[1];j++){
			if((i^j)>mx[2]) continue;
			ans=Plus(ans,1ll*sum[0][i]*sum[1][j]%P*sum[2][i^j]%P);
		}
	printf("%d\n",ans);
	
	return 0;
}