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;

const int64_t MOD = 998244353;
void add(int64_t& a, int64_t b){
    a = (a+b) % MOD;
}
void mul(int64_t& a, int64_t b){
    a = a*b % MOD;
}

template<typename T>
vector<T> compress(vector<T> A){
    sort(A.begin(), A.end());
    A.erase(unique(A.begin(), A.end()), A.end());
    return A;
}

int main(){
    int N;
    cin >> N;

    vector<int64_t> pw18(N+1, 1);
    int64_t B = int64_t(1e18) % MOD;
    for(int i=1; i<=N; i++) pw18[i] = pw18[i-1] * B % MOD;

    auto mex = [](vector<int>& V){
        V = compress(V);
        int sz = V.size();
        for(int i=0; i<sz; i++) if(V[i] != i) return i;
        return sz;
    };
    
    vector<int64_t> sum[3];
    int sz[3];
    for(int k=0; k<3; k++){
        int M;
        cin >> M;
        vector<int> edges[100000];
        while(M--){
            int a, b;
            cin >> a >> b;
            a--; b--;
            if(a > b) swap(a, b);
            edges[a].push_back(b);
        }
        vector<int> G(N);
        for(int i=N-1; i>=0; i--){
            vector<int> V;
            for(int j : edges[i]) V.push_back(G[j]);
            G[i] = mex(V);
        }
        sz[k] = *max_element(G.begin(), G.end()) + 1;
        sum[k].resize(sz[k]);
        for(int i=0; i<N; i++) add(sum[k][G[i]], pw18[i+1]);
    }

    int64_t ans = 0;
    for(int i=0; i<sz[0]; i++) for(int j=0; j<sz[1]; j++){
        int k = i^j;
        if(k < sz[2]) add(ans, sum[0][i] * sum[1][j] % MOD * sum[2][k]);
    }

    cout << ans << endl;
    return 0;
}