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;
#define fi first
#define se second
#define Mp make_pair
#define pb push_back

using ll = long long;
using db = double;
using pii = pair<int, int>;
using vi = vector<int>;
mt19937 mrand(time(0));
ll get(ll r) { return ((ll)mrand() * mrand() % r + r) % r; }
ll get(ll l, ll r) { return get(r - l + 1) + l; }

const int P = 998244353;
int n, m, po[200100], sg[3][200100], f[3][200100];
signed main() {
	scanf("%d", &n);
	ll nw = (ll)(1e18) % P; po[0] = 1;
	for(int i = 1; i <= n; i++) po[i] = po[i - 1] * nw % P;
	for(int o = 0; o < 3; o++) {
		scanf("%d", &m);
		vector<vi> G(n + 1);
		for(int i = 1; i <= m; i++) {
			int x, y; scanf("%d %d", &x, &y);
			x > y ? swap(x, y) : void();
			G[x].pb(y);
		}
		for(int i = n; i; i--) {
			set<int> mex;
			for(int j : G[i])
				mex.insert(sg[o][j]);
			while(mex.count(sg[o][i]))
				sg[o][i]++;
		}
		for(int i = 1; i <= n; i++)
			(f[o][sg[o][i]] += po[i]) %= P;
	}

	ll ans = 0;
	for(int i = 0; i < 500; i++)
		for(int j = 0; j < 500; j++)
			(ans += 1ll * f[0][i] * f[1][j] % P * f[2][i ^ j] % P) %= P;
	printf("%lld\n", ans);
	fprintf(stderr, "time=%.4f\n", (db)clock()/CLOCKS_PER_SEC);
	return 0;
	/* 取模直接除，爆零两行泪
	 * 不开ll见祖宗
	 */
}