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>

#define pb push_back
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define debug(x) cerr << #x << " : " << x << '\n'

using namespace std;

typedef long long ll;
typedef long double ld;
typedef string str;
typedef pair<ll, ll> pll;

const ll Mod = 998244353;
const int N = 1e5 + 10;
const ll Inf = 2242545357980376863LL;
const ll Log = 30;

vector<ll> G[N];

ll n, mex[N], ps[N], cnt[4][N];
set<ll> st;
ll Solve(int idx){
	ll m;
	cin >> m;
	for(int i = 1; i <= n; i++) G[i].clear();
	//memset(mk, 0, sizeof mk);
	ll u, v;
	for(int i = 0; i < m; i++){
		cin >> u >> v;
		G[u].pb(v);
		G[v].pb(u);
	}
	ll ans = 0;
	for(int i = n; i >= 1; i--){
		mex[i] = 0;
		st.clear();
		for(auto adj : G[i]) if(adj > i) st.insert(mex[adj]);
		for(auto x : st) if(x == mex[i]) mex[i] ++;
		(cnt[idx][mex[i]] += ps[i]) %= Mod;
	}
	return ans % Mod;
}
ll mul(ll a, ll b){
	return (a * b) % Mod;
}
int main(){
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	ll X = 716070898;
	ps[0] = 1;
	for(int i = 1; i < N; i++) ps[i] = (ps[i - 1] * X) % Mod;
	cin >> n;
	ll res = 0;
	for(int i = 0; i < 3; i++) Solve(i);
	for(int i = 0; i < 1000; i++){
		for(int j = 0; j < 1000; j++) res += mul(cnt[0][i], mul(cnt[1][j], cnt[2][i ^ j]));

	}
cout << res % Mod << '\n';
	return 0;
}
