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;

#define Fi first
#define Se second
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define szz(x) (int)(x).size()
#define rep(i, n) for(int i=0;i<(n);i++)
typedef tuple<int, int, int> t3;

int N;
int color[3][100010];
vector <int> E[3][100010];
const int mod = 998244353;

ll pw(ll x, ll y = mod - 2) {
	ll res = 1;
	while(y) {
		if(y & 1) res = res * x % mod;
		x = x * x % mod;
		y >>= 1;
	}
	return res;
}

const int L = 1<<17;
void FFT_XOR(ll A[], int rev) {
	for(int len=1;len<L;len<<=1) {
		for(int i=0;i<L;i+=len*2) {
			for(int j=0;j<len;j++) {
				ll x = A[i+j];
				ll y = A[i+j+len];
				A[i+j] = (x + y) % mod;
				A[i+j+len] = (x - y + mod) % mod;
			}
		}
	}
	if(rev) rep(i, L) A[i] = A[i] * pw(L) % mod;
}

ll V[3][L];
ll rp[300010];

int main() {
	ll r = 1;
	rep(i, 18) r = r * 10 % mod;
	rp[0] = 1;
	for(int i=1;i<300010;i++) rp[i] = rp[i-1] * r % mod;
	scanf("%d", &N);
	rep(v, 3) {
		int m; scanf("%d", &m);
		while(m--) {
			int x, y;
			scanf("%d%d", &x, &y);
			E[v][x].pb(y);
			E[v][y].pb(x);
		}
		for(int i=N;i;i--) {
			set <int> x;
			for(int e : E[v][i]) if(e > i) x.insert(color[v][e]);
			for(int j=0;;j++) if(x.find(j) == x.end()) {
				color[v][i] = j;
				break;
			}
		}
		for(int i=1;i<=N;i++) {
			int c = color[v][i];
			V[v][c] = (V[v][c] + rp[i]) % mod;
		}
		FFT_XOR(V[v], 0);
	}
	rep(i, L) {
		V[0][i] = V[0][i] * V[1][i] % mod * V[2][i] % mod;
	}
	FFT_XOR(V[0], 1);

	printf("%lld\n", V[0][0]);
	return 0;
}