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>
const int mod=998244353;
#define Rint register int
#define C(n,m) (n<m?0:1ll*fac[n]*invfac[n-m]%mod*invfac[m]%mod)
using namespace std;
int T,n,m,p,k,dp[3][100010],sg[100010],ans=0,ksm_18[100010];
bool vis[1010];
struct edge{
	int to,link,w;
};
struct graph{
	int head[100010],o;
	edge e[2000010];
	void clear(){
		memset(head,0,sizeof(head));
		o=0;
	}
	void add_edge(int u,int v){
		e[++o].to=v,e[o].link=head[u],head[u]=o;
		e[++o].to=u,e[o].link=head[v],head[v]=o;
	}
}g[3];
int main(){
	scanf("%d",&n);
	ksm_18[0]=1,ksm_18[1]=(1000000000000000000ll)%mod;
	for(int i=2;i<=n;i++) ksm_18[i]=1ll*ksm_18[i-1]*ksm_18[1]%mod; 
	for(int t=0;t<3;t++){
		scanf("%d",&m);
		for(int i=1,u,v;i<=m;i++) scanf("%d%d",&u,&v),g[t].add_edge(u,v);
		for(int u=n;u>=1;u--){
			sg[u]=0;
			for(int i=g[t].head[u];i;i=g[t].e[i].link) if(g[t].e[i].to>u) vis[sg[g[t].e[i].to]]=true;
			while(vis[sg[u]]) sg[u]++;
//			cout<<sg[u]<<" ";
			dp[t][sg[u]]=(dp[t][sg[u]]+ksm_18[u])%mod;
			for(int i=g[t].head[u];i;i=g[t].e[i].link) if(g[t].e[i].to>u) vis[sg[g[t].e[i].to]]=false;
		}
//		cout<<"\n";
	}
	for(int i=0;i<=1000;i++) for(int j=0;j<=1000;j++) ans=(ans+1ll*dp[0][i]*dp[1][j]%mod*dp[2][i^j])%mod;
	printf("%d\n",ans);
}