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 lsb(x) (x & (-x))
#define ll long long
#define ull unsigned long long
#define uint unsigned int


using namespace std;

const int MOD = 998244353;

inline int lgput(int a, int b) {
	int ans = 1;
	while(b > 0) {
		if(b & 1) ans = (1LL * ans * a) % MOD;
		b >>= 1;
		a = (1LL * a * a) % MOD;
	}
	return ans;
}

inline void solve(vector<int> &sum, vector<vector<int>> &g, int n) {
	vector<int> grundy(n + 1, 0);
	for(int i = n; i >= 1; i--) {
		set<int> vals;
		for(auto it : g[i]) {
			if(it > i) {
				vals.insert(grundy[it]);
			}
		}
		int j = 0;
		while(vals.find(j) != vals.end()) {
			j++;
		}
		grundy[i] = j;
	}
	sum.resize(*max_element(grundy.begin(), grundy.end()) + 1);
	for(int i = 1; i <= n; i++) {
		sum[grundy[i]] = (sum[grundy[i]] + lgput(10, 18 * i)) % MOD;
	}
}

int main() {
#ifdef HOME
    ifstream cin("A.in");
    ofstream cout("A.out");
#endif
    int n;
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

   	// dp0[x][y][z] =  0 (poti sa-l iei pe x, y, z)
   	//			    != 0 (nu poti sa-l iei)	
   	// dp0[x][y][z] = grundy(x) ^ grundy(y) ^ grundy(z)
    // grundy(x) <= sqrt(M)

    cin >> n;
    vector<vector<int>> sum(3);
    for(int t = 0; t < 3; t++) {
    	int m; cin >> m;
    	vector<vector<int>> g(n + 1);
    	while(m--) {
    		int x, y; cin >> x >> y;
    		g[x].push_back(y);
    		g[y].push_back(x);
    	}
    	solve(sum[t], g, n);
    }
    int ans = 0;
    for(int a = 0; a < (int)sum[0].size(); a++) {
    	for(int b = 0; b < (int)sum[1].size(); b++) {
    		int c = (a ^ b);
    		if(c < (int)sum[2].size()) {
    			ans = (ans + 1LL * sum[0][a] * (1LL * sum[1][b] * sum[2][c] % MOD)) % MOD;
    		}
    	}
    }
    cout << ans;

    return 0;
}