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 <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define all(x) x.begin(),x.end()
#define cmin(a,b) (a>b?a=b:a)
#define cmax(a,b) (a<b?a=b:a)
#define mem(a,k) memset(a,k,sizeof(a))
#define lop(i,s,t) for(int i=s;i<(t);++i)
#define rep(i,s,t) for(int i=s;i<=(t);++i)
#define dec(i,s,t) for(int i=s;i>=(t);--i)
#define fore(i,v) for(int i=g[v],d=es[i].d;i;i=es[i].nxt,d=es[i].d)
using namespace std;

#ifdef CURIOUSCAT
#define Pr(f,...) fprintf(stderr,f,##__VA_ARGS__),fflush(stderr)
#else
#define Pr(f,...)
#endif

typedef long long ll;

template<typename T>
void read(T &x){
	x=0; 
	char c;
	for(c=getchar();!isdigit(c);c=getchar()); 
	for(;isdigit(c);c=getchar())x=x*10+c-'0'; 
}

const int N=1e6+50,P=998244353;

int mul(int a,int b){ return 1ll*a*b%P; }
int add(int a,int b){ a+=b; return a>=P?a-P:a; }
int sub(int a,int b){ a-=b; return a<0?a+P:a; }
int gcd(int a,int b){ return !b?a:gcd(b,a%b); }

int C(int n,int m) { 
	if (m<0||m>n)return 0;
	if(m<=1)return m==1?n:1;
	return 1ll*C(n/2,m/2)*C(n%2,m%2)%2;
}

int n,pw[N],sg[3][N],ans,m,o[N],w[N],mx[3];
vector<int> g[N];
int main(int argc,char *argv[]){
#ifdef CURIOUSCAT
#endif
	read(n); 	
	{int w=1;rep(i,1,18)w=mul(w,10);
	pw[0]=1;rep(i,1,n)pw[i]=mul(pw[i-1],w);}
	lop(k,0,3){
		rep(i,1,n)g[i].clear();
		read(m); 
		rep(i,1,m){
			int u,v; 
			read(u),read(v); 
			if(u>v)swap(u,v); 
			g[u].pb(v); 
		}
		mem(o,0); 
		dec(v,n,1){
			for(int d:g[v])o[w[d]]=v; 
			int x=0;
			while(o[x]==v)++x;
			w[v]=x;
			sg[k][x]=add(sg[k][x],pw[v]);
			if(w[v])Pr("w[%d]=%d\n",v,w[v]); 
			cmax(mx[k],x); 
		}
		Pr("%d,%d\n",sg[k][0],sg[k][1]); 
	}
	rep(i,0,mx[0])rep(j,0,mx[1])ans=add(ans,mul(sg[0][i],mul(sg[1][j],sg[2][i^j]))); 
	printf("%d\n",ans); 
	return 0;
}