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;
#define int long long
const int MOD = 998244353;

inline int add(int x, int y) {
	return (x + y) % MOD;
}

inline int sub(int x, int y) {
	return (x + MOD - y) % MOD;
}

inline int mul(int x, int y) {
	return (long long)x * y % MOD;
}

inline int mod_pow(int a, int b) {
	int res = 1;
	while (b) {
		if (b & 1) {
			res = mul(res, a);
		}
		a = mul(a, a);
		b >>= 1;
	}
	return res;
}

//xor convolution
void fwht(vector<int>& poly, bool rev = false) {
	int n = (int)poly.size();
	for (int i = 1; i < n; i *= 2) {
		for (int j = 0; j < i; j++) {
			for (int k = 0; k < n; k += i * 2) {
				int u = poly[j + k], v = poly[j + k + i];
				poly[j + k] = add(u, v);
				poly[j + k + i] = sub(u, v);
			}
		}
	}
	if (rev) {
		int inv = mod_pow(n, MOD - 2);
		for (int i = 0; i < n; i++) {
			poly[i] = mul(poly[i], inv);
		}
	}
}

vector<int> mul(vector<int> a, vector<int> b) {
	int sm = (int)a.size() + (int)b.size() - 1, size_ = 1;
	while (size_ < sm) size_ *= 2;
	a.resize(size_, 0), b.resize(size_, 0);
	fwht(a), fwht(b);
	for (int i = 0; i < size_; i++) {
		a[i] = mul(a[i], b[i]);
	}
	fwht(a, true);
	a.resize(sm);
	return a;
}
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N;
	cin >> N;
	vector<vector<int> > g[3];

	for (int i = 0; i < 3; i++) {
		g[i].resize(N);
		int M;
		cin >> M;
		for (int j = 0; j < M; j++) {
			int a, b;
			cin >> a >> b;
			a--;
			b--;
			if (a > b) swap(a, b);
			g[i][a].push_back(b);
		}
	}
	pair<vector<int>, vector<int> > v[3];
	vector<int> X(3 * N + 5);
	X[0] = 1;
	X[1] = ((int)1e18) % MOD;
	for (int i = 2; i < X.size(); i++) {
		X[i] = (X[i - 1] * X[1]) % MOD;
	}
	vector<int> A[3];

	for (int i = 0; i < 3; i++) {
		A[i].resize(N + 1, 0);
		vector<int> mk(N, 0);
		vector<int> G(N, 0);
		for (int j = N - 1; j >= 0; j--) {
			set<int> st;
			for (const int& k : g[i][j]) {
				st.insert(G[k]);
			}
			while (st.count(G[j]) != 0) {
				G[j]++;
			}
			A[i][G[j]] = add(A[i][G[j]], X[j + 1]);
		}
	}
	vector<int> T = mul(A[0], A[1]);
	int res = 0;
	for (int i = 0; i < min(T.size(), A[2].size()); i++) {
		res = add(res, T[i] * A[2][i] % MOD);
	}
	cout << res << endl;
}