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>
#define LL long long
#define db double

using namespace std;
const int N=1e5+10,mod=998244353;
int rd()
{
    int x=0,w=1;char ch=0;
    while(ch<'0'||ch>'9'){if(ch=='-') w=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+(ch^48);ch=getchar();}
    return x*w;
}
void ad(int &x,int y){x+=y,x-=x>=mod?mod:0;}
int fpow(int a,int b){int an=1;while(b){if(b&1) an=1ll*an*a%mod;a=1ll*a*a%mod,b>>=1;}return an;}
int ginv(int a){return fpow(a,mod-2);}
int n,pw[N],vs[N],ti,sb[3][N],ans;
queue<int> q;
struct graph
{
	int to[N],nt[N],hd[N],sg[N],tot;
	void adde(int x,int y){++tot,to[tot]=y,nt[tot]=hd[x],hd[x]=tot;}
	void inii()
	{
		for(int x=n;x;--x)
		{
			++ti;
			for(int i=hd[x];i;i=nt[i])
			{
				int y=to[i];
				vs[sg[y]]=ti;
			}
			while(vs[sg[x]]==ti) ++sg[x];
		}
	}
}gg[3];

int main()
{
	n=rd();
	pw[0]=1,pw[1]=fpow(10,18);
	for(int i=2;i<=n;++i) pw[i]=1ll*pw[i-1]*pw[1]%mod;
	for(int h=0;h<=2;++h)
	{
		int m=rd();
		for(int i=1;i<=m;++i)
		{
			int x=rd(),y=rd();
			if(x>y) swap(x,y);
			gg[h].adde(x,y);
		}
		gg[h].inii();
		for(int i=1;i<=n;++i) ad(sb[h][gg[h].sg[i]],pw[i]);
	}
	for(int i=0;i<=1000;++i)
		for(int j=0;j<=1000;++j)
			ad(ans,1ll*sb[0][i]*sb[1][j]%mod*sb[2][i^j]%mod);
	printf("%d\n",ans);
	return 0;
}
