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 <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#include <complex>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <cassert>
#include <fstream>
#include <utility>
#include <functional>
#include <time.h>
#include <stack>
#include <array>
#define popcount __builtin_popcount
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;

int main()
{
	int n;
	cin>>n;
	const ll MOD=998244353;
	ll w[100010];
	w[0]=1, w[1]=1000000000000000000ll%MOD;
	for(int i=2; i<=n; i++) w[i]=w[i-1]*w[1]%MOD;
	int m[3];
	vector<ll> c[3];
	for(int t=0; t<3; t++){
		cin>>m[t];
		vector<vector<int>> g(n);
		vector<int> v(n);
		for(int i=0; i<m[t]; i++){
			int x, y; cin>>x>>y; x--; y--;
			if(x>y) swap(x, y);
			g[x].push_back(y);
		}
		for(int i=n-1; i>=0; i--){
			set<int> st;
			for(auto y:g[i]){
				st.insert(v[y]);
			}
			for(int k=0; ; k++){
				if(st.find(k)==st.end()){
					v[i]=k;
					break;
				}
			}
		}
		int mx=0;
		for(int i=0; i<n; i++) mx=max(mx, v[i]);
		c[t].resize(mx+1);
		for(int i=0; i<n; i++){
			c[t][v[i]]+=w[i+1];
			if(c[t][v[i]]>=MOD) c[t][v[i]]-=MOD;
		}
	}
	ll ans=0;
	for(int i=0; i<c[0].size(); i++){
		for(int j=0; j<c[1].size(); j++){
			int k=(i^j);
			if(k>=c[2].size()) continue;
			(ans+=c[0][i]*c[1][j]%MOD*c[2][k])%=MOD;
		}
	}
	cout<<ans<<endl;
	return 0;
}