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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;
using ll = long long;
const ll MOD = 998244353;

ll modAdd(ll a, ll b) {
	return a+b >= MOD ? a+b-MOD : a+b;
}
ll modSub(ll a, ll b) {
	return a < b ? a-b+MOD : a-b;
}
ll modPow(ll a, ll b) {
	if (b & 1) return a * modPow(a, b-1) % MOD;
	if (b == 0) return 1;
	return modPow(a*a % MOD, b / 2);
}

// res[k] = \sum_{i,j : i^j == k} a[i] b[j]
// overflows if n res[i] overflows
void xorTF(vector<ll>& v) {
	for (int b = 1; b < v.size(); b *= 2) {
		for (int i = 0; i < v.size(); ++i) {
			if (i & b) continue;
			ll ri = v[i^b];
			v[i^b] = modSub(v[i], ri);
			v[i] = modAdd(v[i], ri);
		}
	}
}
vector<ll> xorConv(vector<ll> a, vector<ll> b) {
	xorTF(a); xorTF(b);
	for (int i = 0; i < a.size(); ++i) a[i] = (a[i] * b[i]) % MOD;
	xorTF(a);

	ll inv = modPow(a.size(), MOD-2);
	for (int i = 0; i < a.size(); ++i) a[i] = (a[i] * inv) % MOD;
	return a;
}

const int N = 1e5;
vector<int> conns[3][N];
int grundy[3][N];
vector<ll> vals[3];

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);

	int n;
	cin >> n;

	for (int j = 0; j < 3; ++j) {
		int m;
		cin >> m;
		for (int i = 0; i < m; ++i) {
			int a, b;
			cin >> a >> b;
			--a; --b;
			conns[j][a].push_back(b);
			conns[j][b].push_back(a);
		}
		for (int i = n-1; i >= 0; --i) {
			vector<int> hits;
			for (auto t : conns[j][i]) {
				if (t > i) hits.push_back(grundy[j][t]);
			}
			sort(hits.begin(), hits.end());

			for (auto v : hits) {
				if (grundy[j][i] == v) ++grundy[j][i];
			}
		}
	}

	int h = 1;
	while(h < n) h <<= 1;
	for (int j = 0; j < 3; ++j) vals[j].resize(h, 0);

	ll base = modPow(10, 18);
	ll cur = base;
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < 3; ++j) {
			vals[j][grundy[j][i]] = modAdd(vals[j][grundy[j][i]], cur);
		}
		cur = (cur * base) % MOD;
	}

	vector<ll> aux = xorConv(vals[0], xorConv(vals[1], vals[2]));
	cout << aux[0] << '\n';
}
