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<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std;
const int mod=998244353;
inline int pl(int a,int b){return (a+=b)>=mod?a-mod:a;}
int kpow(int a,int b)
{
	int s=1;
	for(;b;b>>=1,a=1ll*a*a%mod)
		if(b&1)s=1ll*s*a%mod;
	return s;
}
void fwt(int f[],int opt,int lim)
{
	int x,y;
	for(int len=1;len<lim;len<<=1)
		for(int j=0;j<lim;j+=len<<1)
			for(int k=0;k<len;++k)
			{
				x=f[j+k],y=f[j+k+len];
				f[j+k]=pl(x,y);
				f[j+k+len]=pl(x,mod-y);
			}
	if(opt==-1)
	{
		int inv=kpow(lim,mod-2);
		for(int i=0;i<lim;++i)f[i]=1ll*f[i]*inv%mod;
	}
}
int n,ans,lim;
int po[100010],s[131072];
struct graph
{
	int m;
	vector<int>E[100010];
	int sg[100010];
	int st[100010],tp,a[131072];
	void init()
	{
		scanf("%d",&m);
		for(int i=1,x,y;i<=m;++i)
		{
			scanf("%d %d",&x,&y);
			if(x>y)swap(x,y);
			E[x].emplace_back(y);
		}
		for(int i=n;i;--i)
		{
			tp=0;
			for(auto u:E[i])st[++tp]=sg[u];
			sort(st+1,st+1+tp);
			int cur=-1;
			for(int j=1;j<=tp;++j)
			{
				if(st[j]==cur||st[j]==cur+1)cur=st[j];
				else break;
			}
			sg[i]=cur+1;
			a[sg[i]]=(a[sg[i]]+po[i])%mod;
		}
		fwt(a,1,lim);
	}
}g1,g2,g3;
int main()
{
	po[0]=1;
	int w=1000000000000000000%mod;
	for(int i=1;i<=100000;++i)po[i]=1ll*po[i-1]*w%mod;
	scanf("%d",&n);
	lim=1;
	while(lim<n)lim<<=1;
	g1.init();
	g2.init();
	g3.init();
	for(int i=0;i<lim;++i)s[i]=1ll*g1.a[i]*g2.a[i]%mod*g3.a[i]%mod;
	fwt(s,-1,lim);
	printf("%d",s[0]);
	return 0;
}
