Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < (int)(n); i ++)
#define rep1(i, n) for(int i = 1; i <= (int)(n); i ++)
#define MP make_pair

using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int MOD = 998244353;

int power(int x, int t)
{
	int ret = 1;
	while(t > 0) {
		if(t & 1) ret = 1LL * ret * x % MOD;
		x = 1LL * x * x % MOD;
		t >>= 1;
	}
	return ret;
}

int n, m;
int dp[100005];
bool hv[100005];
vector<int> G[100005];
int cur[131072];
int cnt[131072];

void gen_cnt(int x)
{
	scanf("%d", &m);
	rep1(i, n) G[i].clear();
	rep(i, m) {
		int u, v;
		scanf("%d%d", &u, &v);
		G[min(u, v)].push_back(max(u, v));
	}
	for(int i = n; i >= 1; i --) {
		rep(j, G[i].size()) hv[dp[G[i][j]]] = true;
		for(dp[i] = 0; hv[dp[i]]; dp[i] ++);
		rep(j, G[i].size()) hv[dp[G[i][j]]] = false;
	}
	
	rep(i, 131072) cur[i] = 0;
	rep1(i, n) cur[dp[i]] = (cur[dp[i]] + power(10, 18 * i)) % MOD;
	rep(i, 17) rep(j, 131072) if(!(j >> i & 1)) {
		int ev = cur[j], od = cur[j | 1 << i];
		cur[j] = (ev + od) % MOD;
		cur[j | 1 << i] = (ev - od + MOD) % MOD;
	}
	rep(i, 131072) cnt[i] = 1LL * cnt[i] * cur[i] % MOD;
}

int main()
{
	scanf("%d", &n);
	int ans = 0;
	rep(i, 131072) cnt[i] = 1;
	rep(i, 3) gen_cnt(i);
	rep(i, 131072) ans = (ans + cnt[i]) % MOD;
	ans = 1LL * ans * power(131072, MOD - 2) % MOD;
	printf("%d\n", ans);
	return 0;
}