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
#include <bits/stdc++.h>

using namespace std;

using ll = long long;

mt19937 mt(736);

const int mod = 998244353;


int powmod10(int n)
{
	static vector<int> dp = {1};

	while (n >= dp.size())
		dp.push_back((10ll * dp.back()) % mod);

	return dp[n];
}


void solve(istream &cin = std::cin, ostream &cout = std::cout)
{
	int n;

	cin >> n;

	array<vector<vector<int>>, 3> xyz;

	for (auto &gr : xyz)
	{
		gr.resize(n);

		int m;

		cin >> m;

		for (int i = 0; i < m; i++)
		{
			int a, b;

			cin >> a >> b;
			a--;
			b--;

			gr[a].push_back(b);
			gr[b].push_back(a);
		}
	}

	array<vector<int>, 3> mex;

	array<ll, 3> wm{}, bm{};

	for (int i = 0; i < 3; i++)
	{
		mex[i].resize(n, -1);

		for (int j = n - 1; j >= 0; j--)
		{
			unordered_set<int> neigh;

			for (auto it : xyz[i][j])
				neigh.insert(mex[i][it]);

			for (mex[i][j] = 0; neigh.count(mex[i][j]); mex[i][j]++);
		}
	}

	array<vector<ll>, 3> grundy;

	for (int i = 0; i < 3; i++)
	{
		grundy[i].resize(*max_element(mex[i].begin(), mex[i].end()) + 1);

		for (int j = 0; j < n; j++)
			grundy[i][mex[i][j]] += powmod10(18 * (j + 1));

		for (auto &it : grundy[i])
			it %= mod;
	}

	ll ans = 0;

	for (int x = 0; x < grundy[0].size(); x++)
		for (int y = 0; y < grundy[1].size(); y++)
		{
			auto z = x ^y;

			if (z < grundy[2].size())
				ans += grundy[0][x] * grundy[1][y] % mod * grundy[2][z] % mod;
		}

	cout << ans % mod << endl;
}


int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);

	cout << fixed;

#ifdef LOCAL
	auto st = clock();

	ifstream fin("../input.txt");

	solve(fin);

	cout << "clock: " << (clock() - st) / (double) CLOCKS_PER_SEC << endl;
#else
	solve();
#endif

	return 0;
}
