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 ll long long
#define ld long double
#define db double
#define pint pair<int,int>
#define mk(x,y) make_pair(x,y)
#define fi first
#define se second
#define Rep(x,y,z) for(int x=y;x<=z;x++)
#define Red(x,y,z) for(int x=y;x>=z;x--)
using namespace std;
const int Mod=998244353,Inv2=Mod+1>>1,MAXN=1e5+5,ba=716070898;
char buf[1<<12],*pp1=buf,*pp2=buf,nc;int ny;
inline char gc() {return pp1==pp2&&(pp2=(pp1=buf)+fread(buf,1,1<<12,stdin),pp1==pp2)?EOF:*pp1++;}
//inline char gc(){return getchar();}
inline int read(){
	int x=0;for(ny=1;nc=gc(),(nc<48||nc>57)&&nc!=EOF;)if(nc==45)ny=-1;if(nc<0)return nc;
	for(x=nc-48;nc=gc(),47<nc&&nc<58&&nc!=EOF;x=(x<<3)+(x<<1)+(nc^48));return x*ny;
}
inline int Add(int x,int y){return x+y>=Mod?x+y-Mod:x+y;}
inline int Sub(int x,int y){return x-y<0?x-y+Mod:x-y;}
inline void Div(int&x){if(x&1)x=1ll*x*Inv2%Mod;else x/=2;}
int n,m,len,SG[MAXN],Sum[5][MAXN],ans[MAXN],S[MAXN],N,Fpow[MAXN];
vector<int>G[MAXN];
inline int Get(int x){
	N=0;for(auto y:G[x])S[++N]=SG[y];sort(S+1,S+N+1);
	S[0]=-1;Rep(i,1,N)if(S[i]>S[i-1]+1)return S[i-1]+1;
	return S[N]+1;
}
inline void FwtXor(int*F,int len,int opt){
	for(int i=1;i<len;i<<=1)for(int j=0;j<len;j+=i<<1)for(int k=0;k<i;k++){
		int x=F[i+j+k],y=F[j+k];F[j+k]=Add(x,y),F[i+j+k]=Sub(y,x);if(opt==-1)Div(F[j+k]),Div(F[i+j+k]);
	}
}
int main(){
//	freopen("std.in","r",stdin);
//	freopen("std.out","w",stdout);
	n=read();Fpow[0]=1;Rep(i,1,n)Fpow[i]=1ll*Fpow[i-1]*ba%Mod;
	for(int t=1;t<=3;t++){
		m=read(),len=1024;
		Rep(i,1,m){int x=read(),y=read();if(x<y)swap(x,y);G[y].push_back(x);}
		Red(i,n,1)SG[i]=Get(i),Sum[t][SG[i]]=(Sum[t][SG[i]]+Fpow[i])%Mod;FwtXor(Sum[t],len,1);
		Rep(i,1,n)G[i].clear();
	}for(int i=0;i<len;i++)ans[i]=1ll*Sum[1][i]*Sum[2][i]%Mod*Sum[3][i]%Mod;
	FwtXor(ans,len,-1);cout<<ans[0]<<"\n";
	return 0;
}