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;

const int mod = 998244353;
vector<int> G[3][101010];
int s[3][101010];
int f[3][512];
int M[3];

int buffer[101010];

int _10[202020];

int main()
{
	_10[0] = 1;
	for(int i = 1; i <= 200000; i++)
		_10[i] = 1ll * _10[i-1] * (((long long)1e18) % mod) % mod;
	
	
	int N;
	scanf("%d", &N);
	
	for(int t = 0; t < 3; t++)
	{
		scanf("%d", &M[t]);
		for(int i = 1; i <= M[t]; i++)
		{
			int u, v;
			scanf("%d%d", &u, &v);
			if(u > v)
				swap(u, v);
			G[t][u].push_back(v);
		}
		for(int i = N; i >= 1; i--)
		{
			for(int x: G[t][i])
				buffer[s[t][x]] = 1;
			while(buffer[s[t][i]])
				s[t][i]++;
			for(int x: G[t][i])
				buffer[s[t][x]] = 0;
			f[t][s[t][i]] += _10[i];
			f[t][s[t][i]] %= mod; 
		}
	}
	int ans = 0;
	
	for(int i = 0; i < 512; i++)
	for(int j = 0; j < 512; j++)
	{
		ans += 1ll * f[0][i] * f[1][j] % mod * f[2][i^j] % mod;
		if(ans >= mod)
			ans -= mod;
	}
	printf("%d\n", ans);
	return 0;
}