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<bits/stdc++.h>
using namespace std;
#define maxn 100010
const int p=998244353;
template <typename T>inline T read()
{
    register T sum=0;
    register char cc=getchar();
    int sym=1;
    while(cc!='-'&&(cc>'9'||cc<'0'))cc=getchar();
    if(cc=='-')sym=-1,cc=getchar();
    sum=sum*10+cc-'0';
    cc=getchar();
    while(cc>='0'&&cc<='9')sum=sum*10+cc-'0',cc=getchar();
    return sym*sum;
}
template <typename T>inline T read(T &a)
{
    a=read<T>();
    return a;
}
template <typename T,typename... Others> inline void read(T& a, Others&... b)
{
    a=read(a);
	read(b...);
}
int n,m,mx,ans,SG[3][maxn],sum[3][maxn],buck[maxn];
vector<int>G[3][maxn];
void Add(int &x,int y)
{
	x+=y;
	if(x>=p)
		x-=p;
}
int mul(int x,int y)
{
	return 1ll*x*y%p;
}
int fpow(int x,int y)
{
	int res=1;
	while(y)
	{
		if(y&1)
			res=mul(res,x);
		x=mul(x,x);
		y>>=1;
	}
	return res;
}
int main()
{
	read(n);
	for(int k=0;k<3;k++)
	{
		read(m);
		for(int i=1;i<=m;i++)
		{
			int x,y;
			read(x,y);
			if(x>y)swap(x,y);
			G[k][x].push_back(y);
		}
		for(int i=n;i>=1;i--)
		{
			for(auto j:G[k][i])
				buck[SG[k][j]]=true;
			for(int j=0;j<=mx+1;j++)
			{
				if(buck[j])
					continue;
				SG[k][i]=j;
				mx=max(mx,j);
				break;
			}
			Add(sum[k][SG[k][i]],fpow(10,18*i));
			for(auto j:G[k][i])
				buck[SG[k][j]]=false;
		}
	}
	for(int a=0;a<=mx;a++)
		if(sum[0][a])
			for(int b=0;b<=mx;b++)
				if(sum[1][b])
					Add(ans,mul(mul(sum[0][a],sum[1][b]),sum[2][a^b]));
	printf("%d\n",ans);
    return 0;
}