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;
long long arr[100010];
long long g1[100010]={0,},g2[100010]={0,},g3[100010]={0,};
long long f1[100010]={0,},f2[100010]={0,},f3[100010]={0,};
long long var[100010]={0,};
vector<long long>G1[100010],G2[100010],G3[100010];
long long pw10[100010];
long long mod=998244353;
int main(){
	long long a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z,tmp,ans,cnt,max1=0,max2=0,max3=0;
	scanf("%lld",&n);
	scanf("%lld",&tmp);
	for(i=0;i<tmp;i++){
		scanf("%lld %lld",&a,&b);
		G1[a].push_back(b);
		G1[b].push_back(a);
	}
	scanf("%lld",&tmp);
	for(i=0;i<tmp;i++){
		scanf("%lld %lld",&a,&b);
		G2[a].push_back(b);
		G2[b].push_back(a);
	}
	scanf("%lld",&tmp);
	for(i=0;i<tmp;i++){
		scanf("%lld %lld",&a,&b);
		G3[a].push_back(b);
		G3[b].push_back(a);
	}
	pw10[0]=1;
	z=1000000000000000000%mod;
	for(i=1;i<100010;i++) pw10[i]=(pw10[i-1]*z)%mod;
	for(i=n;i>=1;i--){
		for(j=0; j<G1[i].size(); j++) if(G1[i][j]>i) var[g1[G1[i][j]]]=1;
		for(j=0;j<100010;j++) if(var[j]==0) {
			g1[i]=j;
			if(max1<j) max1=j;
			break;
		}
		for(j=0; j<G1[i].size(); j++) if(G1[i][j]>i) var[g1[G1[i][j]]]=0;
	}
	for(i=n;i>=1;i--){
		for(j=0; j<G2[i].size(); j++) if(G2[i][j]>i) var[g2[G2[i][j]]]=1;
		for(j=0;j<100010;j++) if(var[j]==0) {
			g2[i]=j;
			if(max2<j) max2=j;
			break;
		}
		for(j=0; j<G2[i].size(); j++) if(G2[i][j]>i) var[g2[G2[i][j]]]=0;
	}
	for(i=n;i>=1;i--){
		for(j=0; j<G3[i].size(); j++) if(G3[i][j]>i) var[g3[G3[i][j]]]=1;
		for(j=0;j<100010;j++) if(var[j]==0) {
			g3[i]=j;
			if(max3<j) max3=j;
			break;
		}
		for(j=0; j<G3[i].size(); j++) if(G3[i][j]>i) var[g3[G3[i][j]]]=0;
	}
	//for(i=1;i<=n;i++)printf("%lld ",g1[i]);
	//for(i=1;i<=n;i++)printf("%lld ",g2[i]);
	//for(i=1;i<=n;i++)printf("%lld ",g3[i]);
	for(i=1;i<=n;i++){
		f1[g1[i]]=(f1[g1[i]]+pw10[i])%mod;
		f2[g2[i]]=(f2[g2[i]]+pw10[i])%mod;
		f3[g3[i]]=(f3[g3[i]]+pw10[i])%mod;
	}
	ans=0;
	for(a=0;a<=max1;a++)
	for(b=0;b<=max2;b++)
	for(c=0;c<=max3;c++){
		if(((a^b)^c)==0) ans=(ans+(((f1[a]*f2[b])%mod)*f3[c])%mod)%mod;
	}
	printf("%lld",ans);
}