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;
using ll = long long;

const ll mod = 998244353;
const ll x = 716070898;

vector <int> G[101010];
vector <ll> V[3];
set <int> S1, S2;
ll C[101010], ans;
int n;

int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);

	int m, u, v, i, j, k;
	ll t, s;

	cin >> n;

	for(i = 1, t = x; i <= n; i ++){
		C[i] = t; t = t * x % mod;
	}

	for(i = 0; i < 3; i ++){
		for(j = 1; j <= n; j ++){
			G[j].clear();
			S1.insert(j);
		}
		cin >> m;
		for(; m --; ){
			cin >> u >> v;
			G[u].push_back(v);
			G[v].push_back(u);
		}
		for(; !S1.empty(); ){
			for(s = 0; !S1.empty(); ){
				u = *prev(S1.end()); S1.erase(u);
				s = (s + C[u]) % mod;
				for(int &t: G[u]){
					if(S1.find(t) != S1.end()){
						S1.erase(t);
						S2.insert(t);
					}
				}
			}
			V[i].push_back(s);
			swap(S1, S2);
		}
	}

	for(i = 0; i < V[0].size(); i ++){
		for(j = 0; j < V[1].size(); j ++){
			k = i ^ j;
			if(k < V[2].size()){
				ans = (ans + V[0][i] * V[1][j] % mod * V[2][k]) % mod;
			}
		}
	}

	cout << ans << "\n";

	return 0;
}