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;

typedef long long ll;
const int maxn = 1e5 + 5, mod = 998244353;

inline int gi()
{
	char c = getchar();
	while (c < '0' || c > '9') c = getchar();
	int sum = 0;
	while ('0' <= c && c <= '9') sum = sum * 10 + c - 48, c = getchar();
	return sum;
}

#define eb emplace_back

int n, m, sg[maxn], cnt[3][maxn], mx[3], vis[maxn], pw[maxn];
vector<int> e[maxn];

void solve(int *cnt, int &mx)
{
	m = gi();
	for (int i = 1; i <= n; ++i) e[i].clear();
	for (int u, v, i = 1; i <= m; ++i) {
		u = gi(); v = gi();
		if (u > v) swap(u, v);
		e[u].eb(v);
	}
	fill(vis, vis + n + 1, 0);
	for (int u = n; u >= 1; --u) {
		for (int v : e[u]) vis[sg[v]] = u;
		int mex = 0;
		while (vis[mex] == u) ++mex;
		mx = max(mx, sg[u] = mex);
		cnt[sg[u]] = (cnt[sg[u]] + pw[u]) % mod;
	}
}

int main()
{
	n = gi();
	int base = ((ll)1e18) % mod;
	pw[0] = 1;
	for (int i = 1; i <= n; ++i) pw[i] = (ll)pw[i - 1] * base % mod;
	for (int i = 0; i < 3; ++i) solve(cnt[i], mx[i]);

	int ans = 0;
	for (int i = 0; i <= mx[0]; ++i)
		for (int j = 0; j <= mx[1]; ++j)
			ans = (ans + (ll)cnt[0][i] * cnt[1][j] % mod * cnt[2][i ^ j]) % mod;
	printf("%d\n", ans);
	
	return 0;
}
