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<vector>
#include<cstring>
#include<iostream>
#include<map>
#define RI register int
#define CI const int&
#define VI vector <int>:: iterator
using namespace std;
const int N=100005,mod=998244353,base=(long long)1e18%mod;
int n,x,y,ans,pw[N];
inline void inc(int& x,CI y)
{
	if ((x+=y)>=mod) x-=mod;
}
class Graph
{
	private:
		vector <int> v[N]; int m,sg[N];
		inline int SG(CI x)
		{
			if (~sg[x]) return sg[x]; map <int,bool> ext;
			for (VI to=v[x].begin();to!=v[x].end();++to) ext[SG(*to)]=1;
			while (ext[++sg[x]]); return sg[x];
		}
	public:
		int mx,c[N];
		inline void solve(void)
		{
			RI i; for (scanf("%d",&m),i=1;i<=m;++i)
			if (scanf("%d%d",&x,&y),x<y) v[x].push_back(y); else v[y].push_back(x);
			for (memset(sg,-1,sizeof(sg)),i=1;i<=n;++i) if (!~sg[i]) sg[i]=SG(i);
			for (i=1;i<=n;++i) mx=max(mx,sg[i]),inc(c[sg[i]],pw[i]);
		}
		inline void output(void)
		{
			RI i; printf("%d\n",mx);
			for (i=1;i<=n;++i) printf("%d ",sg[i]); putchar('\n');
			for (i=1;i<=n;++i) printf("%d ",c[i]); putchar('\n');
		}
}G[3];
int main()
{
	RI i,j; for (scanf("%d",&n),pw[0]=i=1;i<=n;++i) pw[i]=1LL*pw[i-1]*base%mod;
	for (i=0;i<3;++i) G[i].solve();
	for (i=0;i<=G[0].mx;++i) for (j=0;j<=G[1].mx;++j)
	inc(ans,1LL*G[0].c[i]*G[1].c[j]%mod*G[2].c[i^j]%mod);
	return printf("%d",ans),0;
}