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<bits/stdc++.h>
using namespace std;

typedef long long LL;

const int N=100000,mod=998244353;
 
int add(int a,int b,int p=mod){return a+b>=p?a+b-p:a+b;}
int sub(int a,int b,int p=mod){return a-b<0?a-b+p:a-b;}
int mul(int a,int b,int p=mod){return (LL)a*b%p;}
void sadd(int &a,int b,int p=mod){a=add(a,b,p);}
void ssub(int &a,int b,int p=mod){a=sub(a,b,p);}
void smul(int &a,int b,int p=mod){a=mul(a,b,p);}

int n,m[3];
struct side{
  int y,next;
}e[N*6+9];
int lin[3][N+9],cs;

void Ins(int id,int x,int y){e[++cs].y=y;e[cs].next=lin[id][x];lin[id][x]=cs;}
void Ins2(int id,int x,int y){Ins(id,x,y);Ins(id,y,x);}

void into(){
  scanf("%d",&n);
  for (int i=0;i<3;++i){
	scanf("%d",&m[i]);
	for (int j=1;j<=m[i];++j){
	  int x,y;
	  scanf("%d%d",&x,&y);
	  Ins2(i,x,y);
	}
  }
}

int pw[N+9];

void Get_pw(){
  int pw18=1;
  for (int i=1;i<=18;++i) smul(pw18,10);
  pw[0]=1;
  for (int i=1;i<=n;++i) pw[i]=mul(pw[i-1],pw18);
}

int sg[3][N+9],b[N+9];
int sum[3][N+9],mx[3];

void Get_sg(int id){
  for (int i=n;i>=1;--i){
	for (int j=lin[id][i];j;j=e[j].next)
	  if (e[j].y>i) b[sg[id][e[j].y]]=1;
    for (;b[sg[id][i]];++sg[id][i]);
	sadd(sum[id][sg[id][i]],pw[i]);
	mx[id]=max(mx[id],sg[id][i]);
	for (int j=lin[id][i];j;j=e[j].next) b[sg[id][e[j].y]]=0;
  }
}

int ans;

void Get_ans(){
  for (int i=0;i<=mx[0];++i)
	for (int j=0;j<=mx[1];++j) sadd(ans,mul(mul(sum[0][i],sum[1][j]),sum[2][i^j]));
}

void work(){
  Get_pw();
  Get_sg(0);
  Get_sg(1);
  Get_sg(2);
  Get_ans();
}

void outo(){
  printf("%d\n",ans);
}

int main(){
  into();
  work();
  outo();
  return 0;
}