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>
using namespace std;

const int maxN = 1e5 + 10, mod = 998244353, lim = 1 << 10;

int n, m;
int ans[maxN + 1], g[maxN + 1], pw[maxN + 1];
int flag[maxN + 1], t[maxN + 1], s[maxN + 1];
bool vis[maxN + 1];

vector<int> nxt[maxN + 1];

inline int ADD(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }

inline int read()
{
	int num = 0, f = 1;
	char ch = getchar();
	while( !isdigit( ch ) ) { if(ch == '-') f = -1; ch = getchar(); }
	while( isdigit( ch ) ) num = (num << 3) + (num << 1) + (ch ^ 48), ch = getchar();
	return num * f;
}

inline void dfs(int u)
{
	vis[u] = true;
	for(int i = 0; i < nxt[u].size(); i++) 
		if(!vis[ nxt[u][i] ]) dfs(nxt[u][i]);
	for(int i = 0; i < nxt[u].size(); i++) flag[ g[ nxt[u][i] ] ] = u;

	g[u] = 0;
	while(flag[ g[u] ] == u) g[u] ++;
}

inline void work()
{
	m = read();
	for(int i = 1; i <= n; i++) vis[i] = false, nxt[i].clear();
	for(int i = 0; i < lim; i++) flag[i] = s[i] = 0;

	for(int i = 1; i <= m; i++)
	{
		int x = read(), y = read();
		if(x > y) swap(x, y);
		nxt[x].push_back(y);
	}

	for(int i = 1; i <= n; i++)
		if(!vis[i]) dfs(i);
	
	for(int i = 1; i <= n; i++) s[ g[i] ] = ADD(s[ g[i] ], pw[i]);

	for(int i = 0; i < lim; i++) t[i] = ans[i], ans[i] = 0;
	for(int i = 0; i < lim; i++)
		for(int j = 0; j < lim; j++)
			ans[i ^ j] = ADD(ans[i ^ j], 1ll * t[i] * s[j] % mod);
}

int main()
{
	int base = 1;
	for(int i = 1; i <= 18; i++) base = 10ll * base % mod;

	n = read();
	pw[0] = 1;
	for(int i = 1; i <= n; i++) pw[i] = 1ll * base * pw[i - 1] % mod;
	
	ans[0] = 1;
	for(int i = 1; i <= 3; i++) work();

	printf("%d", ans[0]);
	return 0;
}