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>
using namespace std;
inline int read(){
	int s=0;char ch=getchar();
	while(!isdigit(ch)) ch=getchar();
	while(isdigit(ch)) s=s*10+ch-'0',ch=getchar();
	return s;
} 
const int N=1e5+5;
typedef long long ll;
ll base=1e18,mod=998244353;
int n;
ll pw[N*3];
vector<int> G[N];
int sg[4][N],mx[4];
int f[4][N];
inline void add(int &x,int y){
	x+=y;if(x>=mod) x-=mod; 
}
int main(){
	n=read();
	pw[0]=1;base%=mod;
	for(int i=1;i<=3*n;i++)  pw[i]=pw[i-1]*base%mod;
	for(int T=1;T<=3;T++){
		int m=read();
		for(int i=1;i<=n;i++) G[i].clear();
		for(int i=1;i<=m;i++){
			int u=read(),v=read();
			if(u>v) swap(u,v);
			G[u].push_back(v);
		}
		set<int> s;
		for(int i=n;i>=1;i--){
			s.clear();
			for(int j:G[i]) s.insert(sg[T][j]);
			while(s.find(sg[T][i])!=s.end()) sg[T][i]++;
			mx[T]=max(mx[T],sg[T][i]);
		}
		for(int i=1;i<=n;i++) add(f[T][sg[T][i]],pw[i]);
	}
	int ans=0;
	for(int i=0;i<=mx[1];i++)
		for(int j=0;j<=mx[2];j++)
			add(ans,1ll*f[1][i]*f[2][j]%mod*f[3][i^j]%mod);
	printf("%d",ans);
	return 0;
}