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>
using namespace std;
const int mod=998244353;
struct edge
{
	int v,nxt;
}e[3][500005];
int n,m[3],h[3][100005],t[3],a[3][1000005],vis[100005],bs,inv2,lim,f[1000005];
vector<int> b[100005];
inline int addmod(int x)
{
	return x>=mod?x-mod:x;
}
inline int submod(int x)
{
	return x<0?x+mod:x;
}
int fpow(int x,int y)
{
	int ans=1;
	while(y)
	{
		if(y&1) ans=1ll*ans*x%mod;
		x=1ll*x*x%mod;
		y/=2;
	}
	return ans;
}
void add(int u,int v,int fl)
{
	e[fl][++t[fl]].v=v;
	e[fl][t[fl]].nxt=h[fl][u];
	h[fl][u]=t[fl];
}
void fwt(int *a)
{
	for(int k=1;k<lim;k<<=1)
		for(int i=0,r=k*2;i<lim;i+=r)
			for(int j=0;j<k;j++)
			{
				int x=a[i+j],y=a[i+j+k];
				a[i+j]=addmod(x+y);
				a[i+j+k]=submod(x-y);
			}
}
void ifwt(int *a)
{
	for(int k=1;k<lim;k<<=1)
		for(int i=0,r=k*2;i<lim;i+=r)
			for(int j=0;j<k;j++)
			{
				int x=a[i+j],y=a[i+j+k];
				a[i+j]=1ll*addmod(x+y)*inv2%mod;
				a[i+j+k]=1ll*submod(x-y)*inv2%mod;
			}
}
void solve(int fl)
{
	for(int i=1;i<=n;i++)
		b[i].clear();
	for(int i=n;i>0;i--)
	{
		int nw=0;
	//	printf("fl=%d,nw=%d,i=%d\n",fl,nw,i);
		for(int j=0;j<(int)b[i].size();j++)
			vis[b[i][j]]=1;
		while(vis[nw]) nw++;
		a[fl][nw]=addmod(a[fl][nw]+fpow(bs,i));
		for(int j=h[fl][i];j;j=e[fl][j].nxt)
		{
			int v=e[fl][j].v;
			b[v].push_back(nw);
		}
		for(int j=0;j<(int)b[i].size();j++)
			vis[b[i][j]]=0;
	}
}
int main()
{
	bs=fpow(10,18);
	inv2=fpow(2,mod-2);
	scanf("%d",&n);
	lim=1;
	while(lim<=n) lim<<=1;
	for(int fl=0;fl<3;fl++)
	{
		scanf("%d",&m[fl]);
		for(int i=1;i<=m[fl];i++)
		{
			int u,v;
			scanf("%d%d",&u,&v);
			add(u,v,fl);
			add(v,u,fl);
		}  
		solve(fl);
		fwt(a[fl]);
	}
	for(int i=0;i<lim;i++)
		f[i]=1ll*a[0][i]*a[1][i]%mod*a[2][i]%mod;
	ifwt(f);
	printf("%d",f[0]);
	return 0;
}