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 <iostream>
#include <vector>
constexpr int P = 998244353 , V = 1000000000000000000 % P, INV2 = (P + 1) / 2;
int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    int n;
    std::cin >> n;
    std::vector<int> f{1};
    auto work = [&]() {
        std::vector<std::vector<int>> e(n);
        std::vector<int> c(n);
        int m;
        std::cin >> m;
        for (int i = 0; i < m; ++i) {
            int u, v;
            std::cin >> u >> v;
            --u;
            --v;
            if (u > v)
                std::swap(u, v);
            e[u].push_back(v);
        }
        for (int i = n - 1; i >= 0; --i) {
            std::vector<bool> vis(e[i].size() + 1);
            for (int v : e[i])
                if (c[v] <= int(e[i].size()))
                    vis[c[v]] = true;
            while (vis[c[i]])
                ++c[i];
        }
        std::vector<int> g;
        for (int i = 0, j = V; i < n; ++i, j = 1ll * j * V % P) {
            if (c[i] >= int(g.size()))
                g.resize(c[i] + 1);
            g[c[i]] = (g[c[i]] + j) % P;
        }
        std::vector<int> h;
        for (int i = 0; i < int(f.size()); ++i) {
            for (int j = 0; j < int(g.size()); ++j) {
                if ((i ^ j) >= int(h.size()))
                    h.resize((i ^ j) + 1);
                h[i ^ j] = (h[i ^ j] + 1ll * f[i] * g[j]) % P;
            }
        }
        f = h;
    };
    work();
    work();
    work();
    std::cout << f[0] << "\n";
    return 0;
}