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"
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
#define int long long
typedef pair<int, int> P;
typedef pair<P, int>PP;
const long long mod = 998244353;
vector<int>G[3][100005];
int G2[3][500];
int grundy[3][100005];
int n;

int modpow(int a, int x) {
	int res = 1;
	while (x) {
		if (x & 1)res = res*a%mod;
		a = a*a%mod;
		x >>= 1;
	}
	return res;
}

signed main() {
	 cin >> n;

	 rep(i, 3) {
		 int m; cin >> m;
		 rep(j, m) {
			 int a, b; cin >> a >> b; a--; b--;
			 if (a > b)swap(a, b);
			 G[i][a].push_back(b);
		 }
		 rep(j, n)grundy[i][j] = -1;
		 for (int j = n - 1; j >= 0; j--) {
			 set<int>S;
			 for (int k : G[i][j])S.insert(grundy[i][k]);
			 int a = 0;
			 while (S.count(a))a++;
			 grundy[i][j] = a;
		 }
		 rep(j, n) {
			 G2[i][grundy[i][j]] += modpow(10, 18*(j + 1));
			 G2[i][grundy[i][j]] %= mod;
		 }
	 }

	int ans = 0;

	rep(i, 500) {
		rep(j, 500) {
			int k = i^j;
			ans += G2[0][i] * G2[1][j] %mod* G2[2][k]%mod;
			ans %= mod;
		}
	}

	cout << ans << endl;
}