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>
using namespace std;
#define REP(i,a,b) for(int i=(a),_ed=(b);i<=_ed;++i)
#define DREP(i,a,b) for(int i=(a),_ed=(b);i>=_ed;--i)
#define mp(x,y) make_pair((x),(y))
#define sz(x) (int)(x).size()
#define pb push_back
typedef long long ll;
typedef pair<int,int> pii;
inline int read(){
    register int x=0,f=1;register char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=0;ch=getchar();}
    while(isdigit(ch)){x=x*10+(ch^'0');ch=getchar();}
    return f?x:-x;
}

const int mod=998244353;
int n,pw[300005];
inline void inc(int& x,int y){x=x+y<mod?x+y:x+y-mod;}

struct Graph{
	int m,sg[100005],vis[100005],val[100005];
	vector<int> E[100005];
	void dfs(int u){
		for(int v:E[u])if(!~sg[v])dfs(v);
		for(int v:E[u])vis[sg[v]]=1;
		sg[u]=0;
		while(vis[sg[u]])++sg[u];
		for(int v:E[u])vis[sg[v]]=0;
	}
	void calSG(){
		memset(sg,-1,sizeof sg);
		REP(i,1,n)if(!~sg[i])dfs(i);
		REP(i,1,n)inc(val[sg[i]],pw[i]);
	}
} G[3];

int main(){
	// freopen("in.in","r",stdin);
	int b=((ll)1e18)%mod;
	REP(i,pw[0]=1,300001)pw[i]=1ll*pw[i-1]*b%mod;
	n=read();
	REP(t,0,2){
		G[t].m=read();
		REP(i,1,G[t].m){
			int u=read(),v=read();
			if(u>v)swap(u,v);
			G[t].E[u].pb(v);
		}
		G[t].calSG();
	}
	int ans=0;
	REP(a,0,sqrt(G[0].m)*2)REP(b,0,sqrt(G[1].m)*2){
		int c=a^b;
		inc(ans,1ll*G[0].val[a]*G[1].val[b]%mod*G[2].val[c]%mod);
	}
	printf("%d\n",ans);
	return 0;
}