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<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
ll mod = 998244353;
vector<int>pat[3][101010];
int gr[101010];
ll p10[555555];
vector<ll>calc(int t, int n)
{
	for (int i = n - 1; i >= 0; i--)
	{
		vector<int>f;
		f.resize(pat[t][i].size() + 3);
		for (int j = 0; j < pat[t][i].size(); j++)
		{
			int v = pat[t][i][j];
			if (gr[v] < f.size())f[gr[v]] = 1;
		}
		for (int j = 0; j < f.size(); j++)
		{
			if (f[j] == 0)
			{
				gr[i] = j;
				break;
			}
		}
	}
	vector<ll>r;
	r.resize(2000);
	for (int i = 0; i < n; i++)r[gr[i]] = (r[gr[i]] + p10[i + 1]) % mod;
	return r;
}
int main()
{
	int num;
	scanf("%d", &num);
	for (int i = 0; i < 3; i++)
	{
		int w;
		scanf("%d", &w);
		for (int j = 0; j < w; j++)
		{
			int za, zb;
			scanf("%d%d", &za, &zb);
			za--, zb--;
			if (za > zb)swap(za, zb);
			pat[i][za].push_back(zb);
		}
	}
	p10[0] = 1;
	for (int i = 0; i < 505050; i++)p10[i + 1] = p10[i] * (1000000000000000000LL % mod) % mod;
	//for (int i = 0; i < 505050; i++)p10[i + 1] = p10[i] * (10LL % mod) % mod;
	vector<ll>v[3];
	for (int i = 0; i < 3; i++)
	{
		vector<ll>z = calc(i, num);
		//for (int k = 0; k < z.size(); k++)printf("%lld ", z[k]); printf("\n");
		v[i] = z;
	}
	ll ans = 0;
	for (int i = 0; i < min(num, 1000); i++)
	{
		for (int j = 0; j < min(num, 1000); j++)
		{
			if ((i^j) < num)
			{
				//printf("%d %d %d\n", i, j, i^j);
				ans = (ans + v[0][i] * v[1][j] % mod*v[2][i^j]) % mod;
			}
		}
	}
	printf("%lld\n", ans);
}