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 pb push_back
#define mp make_pair
#define fi first
#define se second

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

template <typename T> bool chkmax(T &x,T y){return x<y?x=y,true:false;}
template <typename T> bool chkmin(T &x,T y){return x>y?x=y,true:false;}

int readint(){
	int x=0,f=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}

const int cys=998244353;
int n;
ll ans[300000];

ll mod(ll x){return x>=cys?x-cys:x;}

ll qpow(ll x,ll p){
	ll ret=1;
	for(;p;p>>=1,x=x*x%cys) if(p&1) ret=ret*x%cys;
	return ret;
}

struct graph{
	int m;
	int col[100005];
	ll A[300000];
	bool vis[100005];
	vector<int> adj[100005];
	void init(){
		for(int i=n;i>=1;i--){
			for(auto v:adj[i]) vis[col[v]]=1;
			while(vis[col[i]]) col[i]++;
			for(auto v:adj[i]) vis[col[v]]=0;
		}
		for(int i=1;i<=n;i++) A[col[i]]=(A[col[i]]+qpow(10,18*i))%cys;
	}
}g[4];

void fwt(ll *A,int m,int f){
	for(int i=1;i<m;i<<=1){
		for(int j=0;j<m;j+=(i<<1)){
			for(int k=j;k<j+i;k++){
				ll x=A[k],y=A[k+i];
				if(f>0) A[k]=mod(x+y),A[k+i]=mod(x+cys-y);
				else A[k]=(x+y)*((cys+1)/2)%cys,A[k+i]=(x+cys-y)*((cys+1)/2)%cys;
			}
		}
	}
}

int main(){
	n=readint();
	int x,y;
	for(int i=1;i<=3;i++){
		g[i].m=readint();
		for(int j=1;j<=g[i].m;j++){
			x=readint(); y=readint();
			if(x>y) swap(x,y);
			g[i].adj[x].pb(y);
		}
		g[i].init();
	}
	int m;
	for(m=1;m<n;m<<=1);
	for(int i=1;i<=3;i++) fwt(g[i].A,m,1);
	for(int i=0;i<m;i++) ans[i]=g[1].A[i]*g[2].A[i]%cys*g[3].A[i]%cys;
	fwt(ans,m,-1);
	printf("%lld\n",ans[0]);
	return 0;
}