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<bits/stdc++.h>
using namespace std;
const int N=14e4,S=131072,mo=998244353,W=716070898;
int n,mi[N],b[N],i;
inline void FWT(int*a){
	int i,j,k,m;
	for(i=2;i<=S;i<<=1)for(j=0,m=i>>1;j<S;j+=i)for(k=0;k<m;++k){
		int x=a[j+k],y=a[j+k+m];
		a[j+k]=(x+y)%mo;a[j+k+m]=(x+mo-y)%mo;
	}
}
struct graph{
	vector<int>e[N];
	int sg[N],a[N];
	void ini(){
		int i,x,y,m;
		scanf("%d",&m);
		for(i=1;i<=m;++i){
			scanf("%d%d",&x,&y);
			if(x>y)swap(x,y);e[x].push_back(y);
		}
		for(i=n;i;--i){
			static int buc[N];int d=e[i].size();
			memset(buc,0,d+2<<2);
			for(int j:e[i])if(sg[j]<d+2)++buc[sg[j]];
			for(sg[i]=0;buc[sg[i]];++sg[i]);
			a[sg[i]]=(a[sg[i]]+mi[i])%mo;
		}
		FWT(a);
	}
}g[3];
int main(){
	for(i=*mi=1;i<N;++i)mi[i]=1ll*mi[i-1]*W%mo;
	scanf("%d",&n);for(i=0;i<3;++i)g[i].ini();
	memcpy(b,g[0].a,sizeof b);
	for(i=0;i<S;++i)b[i]=1ll*b[i]*g[1].a[i]%mo;
	for(i=0;i<S;++i)b[i]=1ll*b[i]*g[2].a[i]%mo;
	FWT(b);
	printf("%lld\n",998236737ll*b[0]%mo);
}