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>

template <class T>
inline void read(T &res)
{
	char ch;
	while (ch = getchar(), !isdigit(ch));
	res = ch ^ 48;
	while (ch = getchar(), isdigit(ch))
		res = res * 10 + ch - 48;
}

using std::vector;
typedef long long ll;
const int N = 1e5 + 5;
const int mod = 998244353;

vector<int> e[N];
int am, bm, cm, n, m, bas;
int sg[N], vis[N], ans[N], b[N], c[N], ex[N];

template <class T>
inline void CkMax(T &x, T y) {x < y ? x = y : 0;}

inline void add(int &x, int y)
{
	x += y;
	x >= mod ? x -= mod : 0;
}

inline void solve()
{
	for (int i = 1; i <= n; ++i)
		e[i].clear();
	read(m);
	for (int i = 1, u, v; i <= m; ++i)
	{
		read(u); read(v);
		if (u > v)
			std::swap(u, v);
		e[u].push_back(v);
	}
	bm = 0;
	for (int i = n; i >= 1; --i)
	{
		sg[i] = 0;
		for (int j = 0, jm = e[i].size(); j < jm; ++j)
			vis[sg[e[i][j]]] = i;
		while (vis[sg[i]] == i)
			++sg[i];
		add(b[sg[i]], ex[i]);
		CkMax(bm, sg[i]);
	}
	cm = 0;
	for (int i = 0; i <= am; ++i)
	if (ans[i])
		for (int j = 0; j <= bm; ++j)
		if (b[j])
		{
			c[i ^ j] = (1ll * ans[i] * b[j] + c[i ^ j]) % mod;
			CkMax(cm, i ^ j);
		}
	for (int i = 0; i <= am; ++i)
		ans[i] = 0;
	for (int i = 0; i <= bm; ++i)
		b[i] = 0;
	am = cm;
	for (int i = 0; i <= cm; ++i)
		ans[i] = c[i], c[i] = 0;
	for (int i = 0; i <= m; ++i)
		vis[i] = 0; 
}

int main()
{
	read(n);
	bas = (ll)(1e18) % mod;
	ex[0] = 1;
	for (int i = 1; i <= n; ++i)
		ex[i] = 1ll * ex[i - 1] * bas % mod;
	
	am = 0;
	ans[0] = 1;
	solve();
	solve();
	solve();
	printf("%d\n", ans[0]); 
}