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 syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<60;
const double pi=acos(-1);
const double eps=1e-8;
const ll mod=998244353;
const int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};

const ll M=1000;
const ll val=(1000000000000000000)%mod;
int n;

vvi f(){
	vvi g(n);
	int m;
	cin>>m;
	for(int i=0;i<m;i++){
		int u,v;
		cin>>u>>v;
		u--;v--;
		if(u>v) swap(u,v);
		g[u].push_back(v);
	}
	return g;
}

vl ff(vvi g){
	vl a(n),b(M);
	for(int i=n-1;i>=0;i--){
		set<int> st;
		for(auto j:g[i]) st.insert(a[j]);
		for(int j=0;;j++) if(st.find(j)==st.end()){
			a[i]=j;
			break;
		}
	}
	ll t=1;
	for(int i=0;i<n;i++){
		(t*=val)%=mod;
		(b[a[i]]+=t)%=mod;
	}
	return b;
}

int main(){
	cin>>n;
	vvi g1=f(),g2=f(),g3=f();
	vi used(n);
	vl a=ff(g2),b=ff(g3),c;
	while(used!=vi(n,1)){
		vi b(n);
		for(int i=n-1;i>=0;i--) if(!used[i]){
			bool flag=1;
			for(auto j:g1[i]) if(b[j]) flag=0;
			if(flag){
				b[i]++;
				used[i]++;
			}
		}
		ll t=1,s=0;
		for(int i=0;i<n;i++){
			(t*=val)%=mod;
			(s+=t*b[i])%=mod;
		}
		c.push_back(s);
	}
	ll res=0;
	for(int i=0;i<M;i++) for(int j=0;j<M;j++){
		int id=(i^j);
		if(id<c.size()) (res+=c[id]*a[i]%mod*b[j])%=mod;
	}
	cout<<res<<endl;
}