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>
using namespace std;

const int MOD = 998244353;
int poww(int a , int b){int tms = 1; while(b){if(b & 1) tms = 1ll * tms * a % MOD; a = 1ll * a * a % MOD; b >>= 1;} return tms;}
int mex[3][503] , N , M , ans; vector < int > nxt[100003];

void calc(int *arr){
	for(int i = 1 ; i <= N ; ++i) nxt[i].clear();
	cin >> M; for(int i = 1 ; i <= M ; ++i){int p , q; cin >> p >> q; nxt[min(p , q)].push_back(max(p , q));}
	static int tmp[503] , val[100003]; memset(val , 0 , sizeof(val));
	for(int i = N ; i ; --i){
		for(auto t : nxt[i]) tmp[val[t]] = 1;
		while(tmp[val[i]]) ++val[i];
		for(auto t : nxt[i]) tmp[val[t]] = 0;
		arr[val[i]] = (arr[val[i]] + poww(10 , 18 * i)) % MOD;
	}
}

int main(){
	cin >> N; calc(mex[0]); calc(mex[1]); calc(mex[2]);
	for(int i = 0 ; i <= 500 ; ++i)
		for(int j = 0 ; j <= 500 ; ++j)
			ans = (ans + 1ll * mex[0][i] * mex[1][j] % MOD * mex[2][i ^ j]) % MOD;
	cout << ans; return 0;
}
