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 oo 1000000010
#define mod  998244353
const int N = 200010;
const int M = 3 * N;

int n , m , u , v;

int po[M] , mx;

vector< vector< int > > g[3];

vector< int > arr[3];


bool has[N];

void make(vector< vector< int > > &g , vector< int > &v){
	v.resize(n);
	for(int i = n - 1; i >= 0;i--){
		for(int j = 0 ;j <= (int)g[i].size() + 2;j++) has[j] = false;
		for(int j = 0 ;j < (int)g[i].size();j++){
			has[v[g[i][j]]] = true;
		}
		while(has[v[i]]) v[i]++;
		mx = max(mx , v[i]);
	}
}

int dp[4][N];



int main(){
	po[0] = 1;
	po[1] = ((long long)1e18) % mod;
	for(int i = 2;i < M;i++){
		po[i] = (long long)po[i - 1] * po[1] % mod;
	}
	scanf("%d",&n);
	for(int i = 0 ;i < 3;i++){
		g[i].resize(n);
		scanf("%d",&m);
		for(int j = 0 ;j < m;j++){
			scanf("%d%d",&u,&v);
			u--,v--;
			if(u > v) swap(u , v);
			g[i][u].push_back(v);
		}
		make(g[i] , arr[i]);
	}
	dp[3][0] = 1;
	for(int j = 25;j >= 0;j--){
		if(((mx >> j) & 1) == 1){
			mx = (1 << (j + 1)) - 1;
			break;
		}
	}
	for(int i = 2;i >= 0;i--){
		for(int x = 0 ; x <= mx;x++){
			dp[i][x] = 0;
			for(int j = 0 ;j < (int)arr[i].size();j++){
				dp[i][x] += (long long)dp[i + 1][(x ^ arr[i][j])] * po[j + 1] % mod;
				if(dp[i][x] >= mod) 
					dp[i][x] -= mod;
			}
		}
	}
	printf("%d\n",dp[0][0]);
	return 0;
}