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;

typedef long long ll;
typedef pair<int, int> pii;
#define x first
#define y second

#ifdef LOCAL
#include "debug.h"
#endif

const int N = int(1e5) + 10;
const int SQRTN = 500;
const int MOD = 998244353;

int n;
int m[3];
vector<int> adj[3][N];
int g[3][N];
int ans_g[4][SQRTN];
int mod[N];

int main() {
	#ifdef LOCAL
	freopen("in", "r", stdin);
	freopen("out", "w", stdout);
	#endif

	scanf("%d", &n);
	for (int i = 0; i < 3; i++) {
		scanf("%d", &m[i]);
		for (int j = 0; j < m[i]; j++) {
			int u, v;
			scanf("%d %d", &u, &v);
			adj[i][u].push_back(v);
			adj[i][v].push_back(u);
		}
	}

	for (int i = 0; i < 3; i++) {
		for (int j = n; j >= 1; j--) {
			vector<int> gs;
			for (int to : adj[i][j]) if (to > j) gs.push_back(g[i][to]);
			sort(gs.begin(), gs.end());
			gs.resize(unique(gs.begin(), gs.end()) - gs.begin());
			g[i][j] = 0;
			while (g[i][j] < gs.size() && gs[g[i][j]] == g[i][j]) g[i][j]++;
		}
	}

	int p = 1;
	for (int i = 0; i < 18; i++) p = p * 10LL % MOD;
	mod[0] = 1;
	for (int i = 1; i < N; i++) mod[i] = mod[i - 1] * 1LL * p % MOD;

	ans_g[3][0] = 1;
	for (int i = 2; i >= 0; i--) {
		for (int j = 1; j <= n; j++) {
			for (int _g = 0; _g < SQRTN; _g++) {
				int ig = _g ^ g[i][j];
				if (ig < SQRTN) {
					(ans_g[i][_g] += mod[j] * 1LL * ans_g[i + 1][ig] % MOD) %= MOD;
				}
			}
		}
	}

	cout << ans_g[0][0] << endl;

	return 0;
}