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 R register
#define mp make_pair
#define ll long long
#define pii pair<int, int>
using namespace std;
const int N = 110000, mod = 998244353;

int n, m, fa[3][N], lim[3];
ll sum[3][N], f[N], ans;
vector<int> to[N];
set<int> mex;

template <class T>
inline void read(T &x) {
	x = 0;
	char ch = getchar(), w = 0;
	while (!isdigit(ch)) w = (ch == '-'), ch = getchar();
	while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
	x = w ? -x : x;
	return;
}

inline ll quickpow(ll base, ll pw) {
	ll ret = 1;
	while (pw) {
		if (pw & 1) ret = ret * base % mod;
		base = base * base % mod, pw >>= 1;
	}
	return ret;
}

int main() {
	int x, y;
	read(n);
	for (R int i = 0; i <= 2; ++i) {
		read(m);
		for (R int j = 1; j <= n; ++j) to[j].clear();
		for (R int j = 1; j <= m; ++j)
			read(x), read(y), to[min(x, y)].push_back(max(x, y));
		for (R int j = n; j; --j) {
			mex.clear();
			for (auto &v : to[j])
				mex.insert(fa[i][v]);
			while (mex.find(fa[i][j]) != mex.end())
				++fa[i][j];
			lim[i] = max(lim[i], fa[i][j]);
			sum[i][fa[i][j]] = (sum[i][fa[i][j]] + quickpow(10, 18 * j)) % mod;
		}
	}
	for (R int i = 0; i <= lim[1]; ++i)
		for (R int j = 0; j <= lim[2]; ++j)
			f[i ^ j] = (f[i ^ j] + sum[1][i] * sum[2][j]) % mod;
	for (R int i = 0; i <= lim[0]; ++i)
		ans = (ans + sum[0][i] * f[i]) % mod;
	cout << ans << endl;
	return 0;
}