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 <iostream>
#include <vector>

using namespace std;

const int MOD = 998244353;

long long modPow(long long a, long long p){
    if(p == 0) return 1;
    auto res = modPow(a, p/2);
    res = (res*res)%MOD;
    if(p%2) res = (res*a)%MOD;
    return res;
}

vector<int> grundy(int N){
    int M; cin >> M;
    vector<vector<int>> g(N);
    for(int i=0;i<M;i++){
        int a, b; cin >> a >> b;
        --a; --b;
        if(a > b) swap(a, b);
        g[a].push_back(b);
    }
    vector<int> res(N);
    vector<int> gr(N, 0);
    for(int i=N-1;i>=0;i--){
        for(int j=0;j<=g[i].size();j++) gr[j] = 0;
        for(auto& t : g[i]) gr[res[t]] = 1;
        for(int j=0;;j++){
            if(!gr[j]){ res[i] = j; break; }
        }
    }
    return res;
}

int main(){
    int N;
    while(cin >> N){
        vector<long long> res(1024, 0);
        res[0] = 1;
        for(int i=0;i<3;i++){
            auto gr = grundy(N);
            vector<long long> dst(1024, 0);
            vector<long long> next(1024, 0);
            for(int j=0;j<N;j++){
                dst[gr[j]] += modPow(10, 18*(j+1));
                dst[gr[j]] %= MOD;
            }
            for(int j=0;j<1024;j++){
                for(int k=0;k<1024;k++){
                    next[k^j] += res[k] * dst[j];
                    next[k^j] %= MOD;
                }
            }
            res = next;
        }
        cout << res[0] << endl;
    }
}