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;

typedef long long ll;
typedef pair<int, int> ii;

const int N = 1e5+100, SQ=512, M=3, MOD = 998244353; 

int add(ll a, int b){
    return (a + b + 2LL*MOD) % MOD;
}
 
int mul(ll a, int b){
    return (a * b) % MOD;
}

void sadd(int &a, int b){
    a = add(a, b);
}

void smul(int &a, int b){
    a = mul(a, b);
}

int bp(int a, int b){
    if(b == 0)
        return 1;
    int x = bp(a, b/2);
    smul(x, x);
    return b&1 ? mul(x, a) : x;
}

vector<int> g[M][N];
int n, m[M], gr[M][N], cnt[M][SQ], res[SQ];

void calc(vector<int> g[], int gr[]){
	memset(gr, 0, N * sizeof gr[0]);
	for(int v=n-1 ; v>=0 ; v--){
		set<int> st;
		for(auto u : g[v])
			st.insert(gr[u]);
		int mex=0;
		while(st.count(mex))
			mex ++;
		gr[v] = mex;
	}
}

void merge(int a[], int b[], int res[]){
	memset(res, 0, SQ*sizeof res[0]);
	for(int i=0 ; i<SQ ; i++)
		for(int j=0 ; j<SQ ; j++)
			sadd(res[i^j], mul(a[i], b[j]));
}

signed main(){
	ios_base::sync_with_stdio(false); cin.tie(NULL);
	cin >> n;
	for(int i=0 ; i<M ; i++){
		cin >> m[i];
		for(int j=0 ; j<m[i] ; j++){
			int v, u;cin >> v >> u;v --;u --;
			if(v>u)
				swap(v, u);
			g[i][v].push_back(u);
		}
		calc(g[i], gr[i]);
		for(int j=0 ; j<n ; j++)
			sadd(cnt[i][ gr[i][j] ], bp(10, 18*(j+1)));
	}
	merge(cnt[0], cnt[1], res);
	for(int i=0 ; i<SQ ; i++)
		cnt[1][i] = res[i];
	merge(cnt[1], cnt[2], res);
	cout << res[0] << "\n";
}























