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 pb push_back

typedef long long LL;

const int N=100005;
const int MOD=998244353;

int n,val[N],sg[3][N],s[3][N*2],tim[N],tmp[N*2];
std::vector<int> e[N];

int main()
{
	scanf("%d",&n);
	val[1]=1;
	for (int i=1;i<=18;i++) val[1]=(LL)val[1]*10%MOD;
	for (int i=2;i<=n;i++) val[i]=(LL)val[i-1]*val[1]%MOD;
	int mx[3]={0},now=0;
	for (int k=0;k<3;k++)
	{
		int m;scanf("%d",&m);
		for (int i=1;i<=m;i++)
		{
			int x,y;scanf("%d%d",&x,&y);
			e[x].pb(y);e[y].pb(x);
		}
		for (int i=n;i>=1;i--)
		{
			now++;
			for (int to:e[i]) if (to>i) tim[sg[k][to]]=now;
			for (sg[k][i]=0;tim[sg[k][i]]==now;sg[k][i]++);
			mx[k]=std::max(mx[k],sg[k][i]);
			e[i].clear();
		}
	}
	for (int k=0;k<3;k++)
		for (int i=1;i<=n;i++)
			(s[k][sg[k][i]]+=val[i])%=MOD;
	for (int i=0;i<=mx[0];i++)
		for (int j=0;j<=mx[1];j++)
			(tmp[i^j]+=(LL)s[0][i]*s[1][j]%MOD)%=MOD;
	int ans=0;
	for (int i=0;i<=mx[2];i++)
		(ans+=(LL)s[2][i]*tmp[i]%MOD)%=MOD;
	printf("%d\n",ans);
	return 0;
}