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>
#include<algorithm>
using namespace std;
const int N=100005;
const int MOD=998244353;
int n,m;
long long ksm(long long a,long long b)
{
	a%=MOD,b%=(MOD-1);
	long long res=1;
	while(b)
	{
		if(b&1) res=res*a%MOD;
		a=a*a%MOD,b>>=1;
	}
	return res;
}
vector<int>G[N];
int sg[N];
int dfs(int u)
{
	if(sg[u]!=-1) return sg[u];
	static int vis[N];
	for(int v:G[u])
	{
		dfs(v);
		vis[sg[v]]=true;
	}
	for(int i=0;;i++)
		if(!vis[i])
		{
			sg[u]=i;
			break;
		}
	for(int v:G[u])
		vis[sg[v]]=false;
	return sg[u];
}
vector<long long> getsg()
{
	for(int i=1;i<=n;i++)
		G[i].clear();
	scanf("%d",&m);
	for(int i=1;i<=m;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		if(x>y) swap(x,y);
		G[x].push_back(y);
	}
	memset(sg,-1,sizeof(sg));
	for(int i=n;i>=1;i--)
		dfs(i);
	int Max=*max_element(sg+1,sg+n+1);
	vector<long long>res;
	res.resize(Max+1);
	for(int i=1;i<=n;i++)
		res[sg[i]]=(res[sg[i]]+ksm(1e18,i))%MOD;
	return res;
}
int main()
{
	scanf("%d",&n);
	vector<long long>x=getsg(),y=getsg(),z=getsg();
	long long ans=0;
	for(int i=0;i<x.size();i++)
		for(int j=0;j<y.size();j++)
		{
			int k=i^j;
			if(k>=z.size()) continue;
			ans=(ans+x[i]*y[j]%MOD*z[k]%MOD)%MOD;
		}
	printf("%lld",ans);
	return 0;
}