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;

#define int long long

const int N = 1e5 + 7, mod = 998244353, base = 716070898, sz = 1005;
vector<int> g[3][N], m(3), power(N);

int n, grundy[3][N], sum[4][sz]; // sum of powers of each grundy value

void dfs(int idx, int u){
	if(g[idx][u].empty()) {
		grundy[idx][u] = 0;
		return;
	}
	set<int> num;
	for(auto v : g[idx][u]){
		if(grundy[idx][v] == -1) dfs(idx, v);
		num.insert(grundy[idx][v]);
	}
	int mex = 0;
	while(num.count(mex)) ++mex;
	grundy[idx][u] = mex;
}

signed main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	
	cin >> n;
	for(int i = 0; i < 3; ++i){
		cin >> m[i];
		for(int j = 0; j < m[i]; ++j){
			int u, v;
			cin >> u >> v;
			if(u > v) swap(u, v);
			g[i][u].push_back(v);
		}
	}
	power[0] = 1;
	for(int i = 1; i < N; ++i) power[i] = (power[i - 1] * base) % mod;
	memset(grundy, -1, sizeof(grundy));
	for(int i = 0; i < 3; ++i){
		for(int j = 1; j <= n; ++j){
			if(grundy[i][j] == -1) dfs(i, j);
		}
	}
	
	for(int i = 0; i < 3; ++i){
		for(int j = 1; j <= n; ++j){
			(sum[i][grundy[i][j]] += power[j]) %= mod;
		}
	}
	
	for(int i = 0; i < sz; ++i){
		for(int j = 0; j < sz; ++j){
			if((i ^ j) < sz) (sum[3][i ^ j] += sum[0][i] * sum[1][j]) %= mod;
		}
	}
	int ans = 0;
	for(int i = 0; i < sz; ++i) (ans += sum[2][i] * sum[3][i]) %= mod;
	cout << ans;
}