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<bits/stdc++.h>
using namespace std;
#define int long long
const int size=1<<20,mod=998244353;
struct nobe
{
	int to,ne;
}e[400400];
int len=0,hea[400400];
inline void adde(int x,int y)
{
	e[++len]=(nobe){y,hea[x]};
	hea[x]=len;
}
inline void upd(int &x)
{
	x+=x>>31&mod;
}
inline long long poww(long long a,long long b=mod-2)
{
	long long res=1;
	while(b)
	{
		if(b&1) res=res*a%mod;
		b>>=1;
		a=a*a%mod;
	}
	return res;
}
int n,m,f[100100],sum[3][100100],sz[3],g[100100],num,ans,fir,inv=poww(fir=poww(10,18));
signed main()
{
	scanf("%lld",&n);
	int graph=0,i=1,j=1,k=0;
	while(graph<3)
	{
		if(len)
		{
			memset(hea,0,sizeof(hea));
			len=0;
		}
		scanf("%d",&m);
		j=1;
		int x,y;
		while(j<=m)
		{
			scanf("%lld%lld",&x,&y);
			if(x>y) swap(x,y);
			adde(x,y);
			++j;
		}
		int now=poww(fir,n);
		i=n;
		while(i)
		{
			++num;
			k=hea[i];
			while(k)
			{
				g[f[e[k].to]]=num;
				k=e[k].ne;
			}
			f[i]=0;
			while(g[f[i]]==num) ++f[i];
			sz[graph]=max(sz[graph],f[i]);
			sum[graph][f[i]]+=now-mod;
			sum[graph][f[i]]%=mod;
			now=(long long)now*inv%mod;
			--i;
		}
		++graph;
	}
	i=0;
	while(i<=sz[0])
	{
		j=0;
		k=0;
		while(j<=sz[1])
		{
			k=i^j;
			ans+=(long long)sum[0][i]*sum[1][j]%mod*sum[2][k]%mod-mod;
			ans%=mod;
			++j;
		}
		++i;
	}
	ans+=mod;
	ans%=mod;
	printf("%lld\n",ans);
	return 0;
}