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>
using namespace std;

typedef long long ll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<ll,ll> pll;
typedef vector<bool> vb;
const ll oo = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-9;
#define sz(c) ll((c).size())
#define all(c) begin(c), end(c)
#define FOR(i,a,b) for (ll i = (a); i < (b); i++)
#define FORD(i,a,b) for (ll i = (b)-1; i >= (a); i--)
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define xx first
#define yy second
#define TR(X) ({ if(1) cerr << "TR: " << (#X) << " = " << (X) << endl; })

// const ll N = 3e5 + 10, MOD = 998244353*1e9, A = 100;
const ll N = 3e5 + 10, MOD = 998244353, A = ll(1e18) % MOD;
ll Apow[N];

const ll HALF = (MOD+1)/2;

vl transform(vl A, bool inverse = false) {
	ll N = sz(A);
	vl B(N,0);
	for(ll p = 1; p < N; p *= 2){
		FOR(i,0,N)B[i] = (((i&p)?-A[i]:A[i]) + A[i^p]) % MOD;
		if(inverse) FOR(i,0,N) B[i] = (B[i] * HALF)%MOD;
		swap(A,B);
	}
	return A;
}

vl multiply(vl p, vl q) {
	int N = 1;
	while(N < max(sz(p), sz(q))) N *= 2;
	p.resize(N, 0);
	q.resize(N, 0);
	p = transform(p),	q = transform(q);
	vector<ll> pq(N);
	FOR(i,0,N) pq[i] = (p[i] * q[i]) % MOD;
	return transform(pq, true);
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	Apow[0] = 1;
	FOR(i,1,N) Apow[i] = A*Apow[i-1] % MOD;

	ll n; cin >> n;
	
	ll N = 1;
	while (N < n) N *= 2;

	vl u(N);
	u[0] = 1;
	
	FOR(dim,0,3) {
		ll m; cin >> m;
		vvl adj(n);
		while (m--) {
			ll a, b;
			cin >> a >> b;
			a--, b--;
			if (a > b) swap(a,b);
			adj[a].pb(b);
		}
	
		vl col(n);
		FORD(i,0,n) {
			set<ll> s;
			for (ll j: adj[i]) s.insert(col[j]);
			while (s.count(col[i])) col[i]++;
		}

		vl v(N);
		FOR(i,0,n) v[col[i]] = (v[col[i]] + Apow[i+1]) % MOD;
		
		u = multiply(u,v);

		// cerr << u[0] << " " << u[1] << endl;
	}

	ll res = u[0];
	if (res < 0) res += MOD;
	cout << res << endl;
}