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 ll long long
#define maxn 100005 /*rem*/
#define mod 998244353
#define db double
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define pi pair<int, int>
#define fi first
#define se second
using namespace std;
int n;
ll qs = 1; 
ll pw[maxn];
vi eg[maxn];
bool fl[3][1005][1005];
int main() {
	for (int i = 0; i < 18; i++) qs *= 10;
	qs %= mod;
	pw[0] = 1;
	for (int i = 1; i < maxn; i++)
		pw[i] = pw[i - 1] * qs % mod;
	int n;
	cin >> n;
	ll mt = 1;
	vector<ll> u[3];
	for (int s = 0; s < 3; s++) {
		int m;
		cin >> m;
		for (int i = 1; i <= n; i++) eg[i].clear();
		for (int i = 0; i < m; i++) {
			int u, v;
			scanf("%d%d", &u, &v);
			eg[u].pb(v), eg[v].pb(u);
		}
		set<int> x;
		for (int i = 1; i <= n; i++)
			x.insert(i);
		while (1) {
			ll cu = 0;
			if (!x.size()) break;
			set<int> y;
			while (x.size()) {
				int r = *(--x.end());
				cu = (cu + pw[r]) % mod;
				x.erase(r);
				for (auto v : eg[r]) {
					if (x.find(v) != x.end()) {
						y.insert(v);
						x.erase(v);
					}
				}
			} 
			u[s].pb(cu);
			x = y;
		}
	}
	ll ans = 0;
	for (int i = 0; i < u[0].size(); i++)
		for (int j = 0; j < u[1].size(); j++)
			for (int k = 0; k < u[2].size(); k++) {
				if (fl[0][i][j] || fl[1][j][k] || fl[2][k][i]) continue;
				fl[0][i][j] = fl[1][j][k] = fl[2][k][i] = 1;
				ll cur = u[0][i] * u[1][j] % mod * u[2][k] % mod;
				ans = (ans + cur) % mod;
			}
	cout << ans << endl;
	return 0;
}