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
#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
#define re register
#define LL long long
#define MOD 998244353
#define MAXN 100005
using namespace std;
LL n,ans,bs[3*MAXN];
LL rd(){
	LL x=0,tp=1;char c;c=getchar();
	while(c<'0' || c>'9'){if(c=='-')tp=-1;c=getchar();}
	while(c>='0' && c<='9'){x=(x*10+c-'0');c=getchar();}
	return x*tp;
}
struct Graph{
	LL m,sg[MAXN],f[MAXN];
	vector<int>e[MAXN];
	void work(){
		int x,y;
		m=rd();
		for(int i=1;i<=m;i++){
			x=rd(),y=rd();
			if(x>y)swap(x,y);
			e[x].push_back(y);
		}
		for(int i=n;i>=1;i--){
			set<int>s;
			int sz=e[i].size();
			for(int j=0;j<sz;j++)s.insert(sg[e[i][j]]);
			while(s.count(sg[i]))sg[i]++;
			(f[sg[i]]+=bs[i])%=MOD;
			// printf("sg[%d]=%d\n",i,sg[i]);
		}
	}
}g[3];
int main(){
	// freopen("testdata.in","r",stdin);
	// freopen("shit.out","w",stdout);
	n=rd();int Sqr=sqrt(n);
	bs[0]=1;LL fuck=1000000000000000000ll%MOD;
	for(int i=1;i<=3*n;i++)bs[i]=bs[i-1]*fuck%MOD;
	for(int i=0;i<3;i++)g[i].work();
	for(int i=0;i<=500;i++)
		for(int j=0;j<=500;j++)
			(ans+=g[0].f[i]*g[1].f[j]%MOD*g[2].f[i^j]%MOD)%=MOD;
	printf("%lld\n",ans);
	// printf("%.2lf\n",(double)clock()/CLOCKS_PER_SEC);
	return 0;
}