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<cstdio>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<vector>
using namespace std;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0' && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
int stack[20];
inline void write(int x)
{
	if(x<0){putchar('-');x=-x;}
	if(!x){putchar('0');return;}
	int top=0;
	while(x)stack[++top]=x%10,x/=10;
	while(top)putchar(stack[top--]+'0');
}
inline void pr1(int x){write(x),putchar(' ');}
inline void pr2(int x){write(x),puts("");}
vector<int>p[100010];
const int mod=998244353;
inline int ad(int x){return x>=mod?x-mod:x;}
int pw[100010],sum[3][100010],sg[100010],mx[3];
bool v[100010];
int main()
{
    //freopen("a.in","r",stdin);
    //freopen("a.out","w",stdout);
    pw[1]=1;
    for(int i=1;i<=18;i++)pw[1]=1LL*pw[1]*10%mod;
   	int n=read(),ans=0;
   	for(int i=2;i<=n;i++)pw[i]=1LL*pw[i-1]*pw[1]%mod;
	for(int T=0;T<3;T++)
	{
		int m=read();
		for(int i=1;i<=n;i++)p[i].clear(),sg[i]=0;
		for(int i=1;i<=m;i++)
		{
			int x=read(),y=read();if(x>y)swap(x,y);
			p[x].push_back(y);
		}sum[T][0]=ad(sum[T][0]+pw[n]);
		for(int i=n-1;i>=1;i--)
		{
			int now=0;
			for(int j=0;j<p[i].size();j++)v[sg[p[i][j]]]=true;
			while(v[now]==true)now++;mx[T]=max(mx[T],now),sg[i]=now;sum[T][now]=ad(sum[T][now]+pw[i]);
			for(int j=0;j<p[i].size();j++)v[sg[p[i][j]]]=false;
		}
	}
	for(int i=0;i<=mx[0];i++)
	{
		for(int j=0;j<=mx[1];j++)ans=ad(ans+1LL*sum[0][i]*sum[1][j]%mod*sum[2][i^j]%mod);
	}pr2(ans);
    return 0;
}