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<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
int n, m;
long long Mod = 998244353, po[101000];
struct Graph {
	vector<int>E[101000];
	int D[101000], m, v[100010], Mx;
	long long C[100010];
	void Input() {
		int i, a, b, j;
		scanf("%d", &m);
		for (i = 0; i < m; i++) {
			scanf("%d%d", &a, &b);
			if (a > b)swap(a, b);
			E[a].push_back(b);
		}
		for (i = n; i >= 1; i--) {
			for (auto &t : E[i])v[D[t]] = 1;
			for (j = 0; v[j]; j++);
			D[i] = j;
			for (auto &t : E[i])v[D[t]] = 0;
			Mx = max(Mx, D[i]);
			C[D[i]] = (C[D[i]] + po[i]) % Mod;
		}
	}
}A, B, C;
int main() {
	int i;
	long long z = 1e18;
	scanf("%d", &n);
	po[0] = 1;
	for (i = 1; i <= n; i++)po[i] = po[i - 1] * (z%Mod) % Mod;
	A.Input();
	B.Input();
	C.Input();
	long long res = 0;
	for (i = 0; i <= A.Mx; i++) {
		for (int j = 0; j <= B.Mx; j++) {
			res = (res + A.C[i] * B.C[j] % Mod*C.C[i^j]) % Mod;
		}
	}
	printf("%lld\n", res);
}