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<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
#define N 100010
const int mod=998244353;
inline int read(){
	int x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
int qpow(int a,int b){
	int ans=1;
	while(b){
		if(b&1)ans=1LL*ans*a%mod;
		a=1LL*a*a%mod;
		b>>=1;
	}
	return ans;
}
const int NUM=qpow(10,18);
int n,cnt[3][633],pw[N*3],ans,sg[N];
vector<int> G[N];
int get_mex(vector<int> &vec){
	static int vis[N],myh=0;
	++myh;
	for(auto x:vec){
		vis[x]=myh;
	}
	for(int k=0;;++k){
		if(vis[k]^myh)return k;
	}
}
void dfs(int u){
	vector<int> tmp;
	for(auto v:G[u]){
		if(!~sg[v])dfs(v);
		tmp.push_back(sg[v]);
	}
	sg[u]=get_mex(tmp);
}
void Solve(int T){
	int m=read();
	for(int i=1;i<=n;++i){
		G[i].clear();
		sg[i]=-1;
	}
	for(int i=1;i<=m;++i){
		int u=read(),v=read();
		if(u>v)swap(u,v);
		G[u].push_back(v);
	}
	for(int i=1;i<=n;++i){
		if(~sg[i])continue;
		dfs(i);
	}
	for(int i=1;i<=n;++i){
		cnt[T][sg[i]]=(cnt[T][sg[i]]+pw[i])%mod;
	}
}
int main(){
	n=read();
	pw[0]=1;
	for(int i=1;i<=3*n;++i){
		pw[i]=1LL*pw[i-1]*NUM%mod;
	}
	for(int T=0;T<3;++T){
		Solve(T);
	}
	for(int i=0;i<512;++i){
		for(int j=0;j<512;++j){
			int k=i^j;
			ans=(ans+1LL*cnt[0][i]*cnt[1][j]%mod*cnt[2][k])%mod;
		}
	}
	printf("%d\n",ans);
	return 0;
}

