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 MAXN = 1e5 + 5;
const int P = 998244353;
const int Q = 716070898;
const int limit = 1000;
typedef long long ll;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
int n, m, x[MAXN], y[MAXN], z[MAXN];
int bit[MAXN], f[MAXN], g[MAXN], res[MAXN];
vector <int> a[MAXN], b[MAXN], c[MAXN];
void update(int &x, int y) {
	x += y;
	if (x >= P) x -= P;
}
void work(vector <int> *a, int *res) {
	static int vis[MAXN];
	for (int i = n; i >= 1; i--) {
		for (auto x : a[i]) if (x > i) vis[res[x]] = true;
		while (vis[res[i]]) res[i]++;
		for (auto x : a[i]) if (x > i) vis[res[x]] = false;
	}
}
int main() {
	read(n);
	read(m);
	for (int i = 1; i <= m; i++) {
		int x, y; read(x), read(y);
		a[x].push_back(y);
		a[y].push_back(x);
	}
	read(m);
	for (int i = 1; i <= m; i++) {
		int x, y; read(x), read(y);
		b[x].push_back(y);
		b[y].push_back(x);
	}
	read(m);
	for (int i = 1; i <= m; i++) {
		int x, y; read(x), read(y);
		c[x].push_back(y);
		c[y].push_back(x);
	}
	work(a, x);
	work(b, y);
	work(c, z);
	bit[0] = 1;
	for (int i = 1; i <= n; i++)
		bit[i] = 1ll * Q * bit[i - 1] % P;
	for (int i = 1; i <= n; i++) {
		update(f[y[i]], bit[i]);
		update(g[z[i]], bit[i]);
	}
	for (int i = 0; i <= limit; i++) {
		for (int j = 0; j <= limit; j++) {
			int k = i ^ j;
			update(res[i], 1ll * f[j] * g[k] % P);
		}
	}
	int ans = 0;
	for (int i = 1; i <= n; i++)
		update(ans, 1ll * res[x[i]] * bit[i] % P);
	cout << ans << endl;
	return 0;
}
