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;
typedef long long ll;
const int mod = 998244353;
inline int add(int a,int b){a+=b;return a>=mod?a-mod:a;}
inline int sub(int a,int b){a-=b;return a<0?a+mod:a;}
inline int mul(int a,int b){return 1ll*a*b%mod;}
inline int qpow(int a,int b){int ret=1;for(;b;b>>=1,a=mul(a,a))if(b&1)ret=mul(ret,a);return ret;}
const int inv2 = (mod+1)>>1;
/* math */
const int N = 1e5+5;
int a[1<<19],b[1<<19],c[1<<19],d[1<<19];
inline void fwt(int *f,int n,int type){
	for(int step=1;step<n;step<<=1){
		for(int i=0;i<n;i+=step<<1){
			for(int j=0;j<step;j++){
				int x=f[i+j], y=f[i+j+step];
				f[i+j] = add(x,y);
				f[i+j+step] = sub(x,y);
				if(type==0){
					f[i+j] = mul(f[i+j], inv2);
					f[i+j+step] = mul(f[i+j+step], inv2);
				}
			}
		}
	}
}
int n,buc[N],sg[N];
vector<int> s[N];
inline void calc(int n,int *q){
	int m;cin >> m;for(int i=1;i<=n;i++)s[i].clear();
	for(int i=1;i<=m;i++){
		int u,v;scanf("%d%d",&u,&v);
		s[min(u,v)].push_back(max(u,v));
	}
	for(int i=n;i;i--){
		sg[i]=0;
		for(size_t j=0;j<s[i].size();j++){
			int v=s[i][j];
			// cout << i << " -> " << v << endl;
			buc[sg[v]]++;
		}
		while(buc[sg[i]])sg[i]++;
		for(size_t j=0;j<s[i].size();j++){
			int v=s[i][j];
			buc[sg[v]]--;
		}
		// cout << i << " " << sg[i] << endl;
		q[sg[i]]=add(q[sg[i]], qpow(10,18*i));
	}
}

int main()
{
	cin >> n;
	calc(n,a),calc(n,b),calc(n,c);
	fwt(a,1<<19,1);
	fwt(b,1<<19,1);
	fwt(c,1<<19,1);
	for(int i=0;i<1<<19;i++){
		d[i]=mul(a[i],mul(b[i],c[i]));
	}
	fwt(d,1<<19,0);
	cout << d[0] << endl;
}