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 endl '\n'
#define fi first
#define se second
#define MOD(n,k) ( ( ((n) % (k)) + (k) ) % (k))
#define forn(i,n) for (int i = 0; i < (n); i++)
#define forr(i,a,b) for (int i = (a); i <= (b); i++)
#define all(v) v.begin(), v.end()
#define pb push_back
 
using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<ii> vii;

const int MX = 100005, mod = 998244353, SQ = (1 << 9);
int n, m, inv2, f[MX];
ll a[3][SQ], res;
vi adj[MX];

ll pot (ll b, int p) {
	ll res = 1;
	while (p) {
		if (p & 1) (res *= b) %= mod;
		(b *= b) %= mod;
		p /= 2;
	}
	return res;
}

void dfs (int u) {
	if (f[u] != -1) return;

	set<int> st;
	for (int v : adj[u])
		if (v > u) {
			dfs(v);
			st.insert(f[v]);
		}

	for (int i = 0;; i++)
		if (!st.count(i)) {
			f[u] = i;
			break;
		}
}

int main () {
	ios_base::sync_with_stdio(0); cin.tie(0);
	inv2 = pot(2, mod - 2);

	cin >> n;
	forn (i, 3) {
		for (int j = 1; j <= n; j++)
			adj[j].clear();

		cin >> m;
		while (m--) {
			int a, b;
			cin >> a >> b;
			adj[a].pb(b);
			adj[b].pb(a);
		}

		memset(f, -1, sizeof(f));
		for (int j = 1; j<= n; j++) {
			dfs(j);
			(a[i][f[j]] += pot(10, 18 * j)) %= mod;
		}
	}

	for (int i = 0; i < SQ; i++)
		for (int j = 0; j < SQ; j++)
			(res += a[0][i] * a[1][j] % mod * a[2][i ^ j]) %= mod;

	cout << res << endl;

	return 0;
}