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>
#define mset(a, b) memset(a, b, sizeof(a))
#define mcpy(a, b) memcpy(a, b, sizeof(a))
using namespace std;
typedef long long LL;
const int MAXN = 100005;

template <typename T> inline void read(T &AKNOI) {
	T x = 0, flag = 1;
	char ch = getchar();
	while (!isdigit(ch)) {
		if (ch == '-') flag = -1;
		ch = getchar();
	}
	while (isdigit(ch)) {
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	AKNOI = flag * x;
}

namespace ModCalculator {
	const int MOD = 998244353;
	inline void Inc(int &x, int y) {
		x += y; if (x >= MOD) x -= MOD;
	}
	inline void Dec(int &x, int y) {
		x -= y; if (x < 0) x += MOD;
	}
	inline int Add(int x, int y) {
		Inc(x, y); return x;
	}
	inline int Sub(int x, int y) {
		Dec(x, y); return x;
	}
	inline int Mul(int x, int y) {
		return 1LL * x * y % MOD;
	}
}
using namespace ModCalculator;
 
inline int ksm(int x, int k) {
    int ret = 1;
   	for (; k; k >>= 1) {
        if (k & 1) ret = Mul(ret, x);
        x = Mul(x, x);
    }
    return ret;
}

int n, tim, vis[MAXN], pw[MAXN];

struct Graph {
	vector<int> e[MAXN];
	int m, sg[MAXN], mxsg, sum[MAXN];
	
	void build() {
		read(m);
		for (int i = 1, u, v; i <= m; ++i) {
			read(u); read(v);
			e[min(u, v)].push_back(max(u, v));
		}
		for (int u = n; u >= 1; --u) {
			++tim;
			for (auto v : e[u]) {
				vis[sg[v]] = tim;
			}
			while (vis[sg[u]] == tim) {
				++sg[u];
			}
			Inc(sum[sg[u]], pw[u]);
			mxsg = max(mxsg, sg[u]);
		}
	}
} g1, g2, g3;

void init() {
	read(n);
	int p18 = ksm(10, 18);
	pw[0] = 1;
	for (int i = 1; i <= n; ++i) {
		pw[i] = Mul(pw[i - 1], p18);
	}
	g1.build();
	g2.build();
	g3.build();
}

void solve() {
	int ans = 0;
	for (int i = 0; i <= g1.mxsg; ++i) {
		for (int j = 0; j <= g2.mxsg; ++j) {
			Inc(ans, Mul(Mul(g1.sum[i], g2.sum[j]), g3.sum[i ^ j]));
		}
	}
	printf("%d\n", ans);
}

int main() {
	init();
	solve();
	return 0;
}
