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>
#define PB push_back
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
using namespace std;
typedef vector<int>VI;
typedef long long LL;
const int N=100005,P=998244353,B=(LL)1e18%P;
int n,pw[N],m[3],sg[3][N],mx[3],f[3][N];VI G[N];bool ban[N];
int main(){
	scanf("%d",&n);
	pw[0]=1;rep(i,1,N-1)pw[i]=1LL*pw[i-1]*B%P;
	rep(i,0,2){
		scanf("%d",&m[i]);
		rep(j,1,n)G[j].clear();
		rep(j,1,m[i]){
			int k1,k2;scanf("%d%d",&k1,&k2);
			if(k2<k1)swap(k1,k2);
			G[k1].PB(k2);
		}
		per(k1,n,1){
			for(auto j:G[k1])ban[sg[i][j]]=1;
			int&k=sg[i][k1];
			while(ban[k])++k;
			for(auto j:G[k1])ban[sg[i][j]]=0;
		}
		rep(j,1,n){
			(f[i][sg[i][j]]+=pw[j])%=P;
			mx[i]=max(mx[i],sg[i][j]);
		}
	}
	int ans=0;
	rep(i,0,mx[0])rep(j,0,mx[1])(ans+=1LL*f[0][i]*f[1][j]%P*f[2][i^j]%P)%=P;
	printf("%d\n",ans);
	return 0;
}