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;

typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;

const int MAXN = 100000;
const ll MOD = 998244353;
const ll PO = 1000000000000000000LL%MOD;
const int SQ = 450;

int n;
vector<int> lis[3][100100];
ll an1[100100];
ll anv[SQ+10];
int an2[100100];
ll anv2[SQ+10];
int an3[100100];
int ma[SQ+10][SQ+10];
ll po[300100];

int chk[100100];
int chk2[100100][SQ+10];
int main(){
	scanf("%d",&n);
	int m;
	for (int t=0;t<3;t++) {
		scanf("%d", &m);
		for (int i = 0; i < m; i++) {
			int a, b;
			scanf("%d%d", &a, &b); a--; b--;
			if (a>b) swap(a,b);
			lis[t][a].push_back(b);
		}
	}
	po[0] = 1;
	for (int i=1;i<=3*n;i++) po[i] = po[i-1]*PO%MOD;
	for (int i=n-1;i>=0;i--) {
		for (int &j : lis[0][i]) chk[an1[j]] = 1;
		int t;
		for (t=0;chk[t];t++);
		an1[i] = t;
		anv[t] = (anv[t]+po[i])%MOD;
		for (int &j : lis[0][i]) chk[an1[j]] = 0;
	}
	for (int i = n - 1; i >= 0; i--) {
		for (int &k : lis[1][i]) chk2[i][an2[k]]=1;
		int t;
		for (t = 0; chk2[i][t]; t++);
		an2[i] = t;
		//anv2[0] = (anv2[0] + anv[t]*po[i]) % MOD;
		for (int &k : lis[1][i]) chk2[i][an2[k]]=0;
	}
	for (int j=0;j<SQ;j++) {
		for (int i=0;i<SQ;i++) {
			for (int k=0;k<i;k++) chk[ma[j][k]]=1;
			for (int k=0;k<j;k++) chk[ma[k][i]]=1;
			int t;
			for (t=0;t<SQ&&chk[t];t++);
			ma[j][i] = t;
			for (int k=0;k<i;k++) chk[ma[j][k]]=0;
			for (int k=0;k<j;k++) chk[ma[k][i]]=0;
		}
		for (int i=0;i<n;i++) {
			anv2[j] = (anv2[j]+anv[ma[j][an2[i]]]*po[i])%MOD;
		}
	}
	ll ans = 0;
	for (int i=n-1;i>=0;i--) {
		for (int &j : lis[2][i]) chk[an3[j]] = 1;
		int t;
		for (t=0;chk[t];t++);
		an3[i] = t;
		ans = (ans+anv2[t]*po[i])%MOD;
		for (int &j : lis[2][i]) chk[an3[j]] = 0;
	}
	printf("%lld\n",ans*po[3]%MOD);

	return 0;
}