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:
/* cerberus97 - Hanit Banga */

#include <iostream>
#include <iomanip>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>
#include <numeric>

using namespace std;

#define pb push_back
#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL)

typedef long long ll;
typedef long double ld;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;

const int N = 1e5 + 10, mod = 998244353;
const ll base = ll(1e18) % mod;

int add(int a, int b) {
	a += b;
	if (a >= mod) a -= mod;
	return a;
}

int mul(ll a, int b) {
	a *= b;
	return a % mod;
}

int pbase[N];

vector<pii> get(int n);

int main() {
	fast_cin();
	int n; cin >> n;
	pbase[0] = 1;
	for (int i = 1; i <= n; ++i) {
		pbase[i] = mul(pbase[i - 1], base);
	}
	vector<vector<pii>> g(3);
	for (auto& v : g) {
		v = get(n);
	}
	vector<int> sum(2 * n + 1, 0);
	for (auto& p1 : g[0]) {
		for (auto& p2 : g[1]) {
			sum[p1.first ^ p2.first] = add(sum[p1.first ^ p2.first], mul(p1.second, p2.second));
		}
	}
	int ans = 0;
	for (auto& p : g[2]) {
		ans = add(ans, mul(p.second, sum[p.first]));
	}
	cout << ans << endl;
}

vector<pii> get(int n) {
	vector<vector<int>> g(n + 1);
	int m; cin >> m;
	while (m--) {
		int u, v;
		cin >> u >> v;
		g[min(u, v)].pb(max(u, v));
	}
	vector<int> v(n + 1, 0);
	map<int, int> temp;
	for (int i = n; i >= 1; --i) {
		set<int> seen;
		for (auto& j : g[i]) {
			seen.insert(v[j]);
		}
		while (seen.count(v[i])) {
			++v[i];
		}
		temp[v[i]] = add(temp[v[i]], pbase[i]);
	}
	vector<pii> ans(temp.begin(), temp.end());
	return ans;
}