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:
#pragma GCC optimize ("Ofast")
#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i=(a); i<(int)(b); i++)
#define FORD(i, a, b) for (int i=a; i>(int)(b); i--)
#define PPC(x) __builtin_popcount(x)
#define SZ(x) ((int)(x).size())
#define MSB(x) (31 - __builtin_clz(x))
#define pb push_back
#define ALL(x) (x).begin(), (x).end()
#define ithBit(m, i) ((m) >> (i) & 1)
#define ft first
#define sd second
#ifdef DEBUG
#include "debug.h"
#else
#define dbg(...) 0
#endif
using namespace std;
 
const int maxN = 1 << 19, maxS = 513, mod = 998244353;
const long long INF = 1000000000000000000ll;

template <class T1, class T2> inline void addMod(T1& a, T2 b)	{   a = (a + b) % mod;  }
template <class T1, class T2> inline void multMod(T1& a, T2 b)	{   a = a * b % mod;    }

long long pows[maxN], T[3][maxS];
vector <int> graph[3][maxN];
int mex[maxN], vis[maxN];
 
int main()
{
	int n;
	scanf ("%d", &n);
	pows[1] = INF % mod;
	FOR(i, 2, n+1)
		pows[i] = pows[i-1] * pows[1] % mod;
	FOR(s, 0, 3)
	{
		vector <int>* G = graph[s];
		int m;
		scanf ("%d", &m);
		while (m--)
		{
			int a, b;
			scanf ("%d%d", &a, &b);
			if (a > b)
				swap(a, b);
			G[a].pb(b);
		}
		
		FORD(v, n, 0)
		{
			for (int u : G[v])
				vis[mex[u]] = v;
			for (mex[v] = 0; vis[mex[v]] == v; mex[v]++) ;
			addMod(T[s][mex[v]], pows[v]);
		}
		
		fill(vis, vis+n+1, 0);
	}
	
	long long res = 0;
	FOR(a, 0, maxS) FOR(b, 0, maxS) FOR(c, 0, maxS) if ((a^b^c) == 0)
	{
		long long tak = T[0][a];
		multMod(tak, T[1][b]);
		multMod(tak, T[2][c]);
		addMod(res, tak);
	}
	
	printf("%lld\n", res);
	return 0;
}