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<iostream>
#include<iomanip>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<map>
using namespace std;
const int maxn=1024;
const int mod=998244353;
int n,m,sg[200005];
long long f[3][2005],pw[100005],h[2005],B=1e18;
vector<int> G[100005];
long long ksm(long long x,long long n)
{
	long long ans=1;
	while(n)
	{
		if(n&1) ans=ans*x%mod;
		x=x*x%mod;
		n>>=1;
	}
	return ans;
}
void FWT(long long a[],int n,int ty)
{
	for(int i=1;i<n;i<<=1)
	for(int j=0;j<n;j++)
	{
		if(j&i)
		{
			long long t0=a[j-i],t1=a[j];
			a[j-i]=(t0+t1)%mod;a[j]=(t0-t1+mod)%mod;
		}
	}
	if(ty==-1)
	{
		long long rev=ksm(n,mod-2);
		for(int i=0;i<n;i++) a[i]=a[i]*rev%mod;
	}
}
int dfs(int x)
{
	if(sg[x]!=-1) return sg[x];
	map<int,bool> g;
	int sz=G[x].size();
	for(int i=0;i<sz;i++) g[dfs(G[x][i])]=1;
	int now=0;
	while(g[now]) now++;
	return sg[x]=now;
}
int main()
{
	B%=mod;
	scanf("%d",&n);
	pw[0]=1;
	for(int i=1;i<=n;i++) pw[i]=pw[i-1]*B%mod;
	for(int t=0;t<3;t++)
	{
		scanf("%d",&m);
		memset(sg,-1,sizeof(sg));
		for(int i=1;i<=n;i++) G[i].clear();
		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);
		}
		for(int i=1;i<=n;i++) if(sg[i]==-1) dfs(i);
		for(int i=1;i<=n;i++) f[t][sg[i]]=(f[t][sg[i]]+pw[i])%mod;
		FWT(f[t],maxn,1);
	}
	for(int i=0;i<maxn;i++) h[i]=1;
	for(int t=0;t<3;t++)
	for(int i=0;i<maxn;i++) h[i]=h[i]*f[t][i]%mod;
	FWT(h,maxn,-1);
	printf("%lld\n",(h[0]%mod+mod)%mod);
}