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>

#define itn int
#define all(x) (x).begin(), (x).end()
#define make_unique(x) sort(all((x))); (x).resize(unique(all((x))) - (x).begin())

using namespace std;

inline int nxt() {
	int x;
	scanf("%d", &x);
	return x;
}

const int mod = 998244353;
const int N = 1 << 17;
long long dp[3][N];

void wh(long long* b, long long* e) {
	if (e == b + 1) {
		return;
	}
	long long *m = b + (e - b) / 2;
	wh(b, m);
	wh(m, e);
	while (m != e) {
		auto x = *b, y = *m;
		*b = x + y;
		*m = x - y;
		++b, ++m;
	}
}

int mex(vector<int> a) {
	make_unique(a);
	int res = 0;
	while (res < (int)a.size() && a[res] == res) {
		++res;
	}
	return res;
}

long long inv(long long a, long long b) {
	return 1 < a ? b - inv(b % a, a) * b / a : 1;
}

int main() {
	int n = nxt();
	const long long rem = (long long)(1e18) % mod;
	vector<long long> deg(n + 1);
	deg[0] = 1;
	for (int i = 1; i <= n; ++i) {
		deg[i] = deg[i - 1] * rem % mod;
	}
	for (int i = 0; i < 3; ++i) {
		int m = nxt();
		vector<vector<int>> a(n);
		for (int j = 0; j < m; ++j) {
			int u = nxt() - 1, v = nxt() - 1;
			if (u > v) {
				swap(u, v);
			}
			a[u].push_back(v);
		}
		vector<int> gr(n);
		for (int j = n - 1; j >= 0; --j) {
			vector<int> grs;
			for (int x : a[j]) {
				grs.push_back(gr[x]);
			}
			gr[j] = mex(grs);
			dp[i][gr[j]] += deg[j + 1];
		}
		for (int j = 0; j < n; ++j) {
			dp[i][j] %= mod;
		}
	}

	for (int i = 0; i < 3; ++i) {
		wh(dp[i], dp[i] + N);
		for (int j = 0; j < N; ++j) {
			dp[i][j] %= mod;
			if (dp[i][j] < 0) {
				dp[i][j] += mod;
			}
		}
	}
	for (int i = 0; i < N; ++i) {
		dp[0][i] *= dp[1][i];
		dp[0][i] %= mod;
		dp[0][i] *= dp[2][i];
		dp[0][i] %= mod;
	}
	wh(dp[0], dp[0] + N);
	dp[0][0] %= mod;
	if (dp[0][0] < 0) {
		dp[0][0] += mod;
	}
	cout << dp[0][0] * inv(N, mod) % mod << "\n";

	return 0;
}