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<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
ll x,kp,y,n,sg[3][100005],inv2=499122177,qz[100005],mr,ans=0;
const ll mx=1<<17,m=17;
ll aq[mx],a[2][mx],ok[mx];
const ll mod=998244353;
vector<int>s[3][100005];
ll M(ll x){return x>=mod?x-mod:x;}
void solve(int ty)
{
	int vis[100005]={0},v;
	scanf("%lld",&kp);
	for(int i=1;i<=kp;i++)
	{
		scanf("%lld%lld",&x,&y);
		if(x>y)swap(x,y);
		s[ty][x].push_back(y);
	}
	for(int i=n-1;i>0;i--)
	{
		for(int j=0;j<s[ty][i].size();j++)
		{
			v=s[ty][i][j];
			vis[sg[ty][v]]=i;
		}
		for(sg[ty][i]=0;vis[sg[ty][i]]==i;sg[ty][i]++);
	}
}
int main()
{
	qz[0]=1;
	mr=1000000000000000000%mod;
	scanf("%lld",&n);
	for(int i=1;i<=n;i++)qz[i]=qz[i-1]*mr%mod;
	solve(0);
	solve(1);
	solve(2);ok[0]=0;
	for(int i=1;i<mx;i++)
	{
		for(int j=0;j<m;j++)
		{
			if((1<<j)&i)
			{
				ok[i]=1-ok[i-(1<<j)];
				break;
			}
		}
	}
	for(int i=1;i<=n;i++)
	{
		aq[sg[0][i]]=(aq[sg[0][i]]+qz[i])%mod;
		a[0][sg[1][i]]=(a[0][sg[1][i]]+qz[i])%mod;
		a[1][sg[2][i]]=(a[1][sg[2][i]]+qz[i])%mod;
	}
	for(int i=0;i<mx;i++)if(ok[i]==1)a[0][i]=M(mod-a[0][i]),a[1][i]=M(mod-a[1][i]);
	for(int i=0;i<m;i++)
	{
		for(int j=0;j<mx;j++)
		{
			if(j&(1<<i))
			{
				x=a[0][j];y=a[0][j-(1<<i)];
				a[0][j]=M(a[0][j]+y);
				a[0][j-(1<<i)]=M(a[0][j-(1<<i)]+mod-x);
				x=a[1][j];y=a[1][j-(1<<i)];
				a[1][j]=M(a[1][j]+y);
				a[1][j-(1<<i)]=M(a[1][j-(1<<i)]+mod-x);
			}
		}
	}
	for(int j=0;j<mx;j++)a[0][j]=a[0][j]*a[1][j]%mod;
	for(int i=0;i<m;i++)
	{
		for(int j=0;j<mx;j++)
		{
			if(j&(1<<i))
			{
				x=a[0][j];y=a[0][j-(1<<i)];
				a[0][j]=(x-y+mod)*inv2%mod;
				a[0][j-(1<<i)]=(x+y)*inv2%mod;
			}
		}
	}
	for(int i=0;i<mx;i++)if(ok[i]==1)a[0][i]=M(mod-a[0][i]);
	for(int i=0;i<mx;i++)ans=(ans+aq[i]*a[0][i])%mod;
	printf("%lld\n",ans);
	return 0;
} 