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>
using namespace std;

const int maxn=100005;
const int mod=998244353;
const int base=716070898;
const int inv2=(mod+1)/2;
int n,m,sz;
vector<int>edge[maxn];
int pw[maxn],dp[maxn],F[4][maxn*4];
bool occ[maxn];

void fwt(int *x,int sz,int tp){
	for(int m=2;m<=sz;m<<=1){
		int mm=m>>1;
		for(int i=0;i<sz;i+=m){
			int *p1=x+i,*p2=x+i+mm;
			for(int j=0;j<mm;j++,p1++,p2++){
				int u=*p1,v=*p2;
				*p1=u+v>=mod?u+v-mod:u+v;
				*p2=u-v<0?u-v+mod:u-v;
				if(tp<0){
					*p1=1LL*(*p1)*inv2%mod;
					*p2=1LL*(*p2)*inv2%mod;
				}
			}
		}
	}
}

int main(){
	scanf("%d",&n);
	pw[0]=1;
	for(int i=1;i<=n;i++)pw[i]=1LL*pw[i-1]*base%mod;
	for(sz=1;sz<n;sz<<=1);
	for(int t=1;t<=3;t++){
		for(int i=1;i<=n;i++){
			dp[i]=0;
			edge[i].clear();
		}
		scanf("%d",&m);
		for(int i=1;i<=m;i++){
			int u,v;
			scanf("%d%d",&u,&v);
			edge[u].push_back(v);
			edge[v].push_back(u);
		}
		for(int i=n;i>=1;i--){
			for(int j=0;j<int(edge[i].size());j++)
				if(edge[i][j]>i)occ[dp[edge[i][j]]]=true;
			while(occ[dp[i]])dp[i]++;
			F[t][dp[i]]+=pw[i];
			if(F[t][dp[i]]>=mod)F[t][dp[i]]-=mod;
			for(int j=0;j<int(edge[i].size());j++)
				if(edge[i][j]>i)occ[dp[edge[i][j]]]=false;
		}
		fwt(F[t],sz,1);
	}
	for(int i=0;i<sz;i++)F[0][i]=1LL*F[1][i]*F[2][i]%mod*F[3][i]%mod;
	fwt(F[0],sz,-1);
	printf("%d\n",F[0][0]);
	return 0;
}