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>
#include <algorithm>
#include <numeric>
#include <functional>
#include <queue>
#include <stack>
#include <bitset>
#include <set>
#include <chrono>
#include <string>

using namespace std;

#define endl '\n'

const int mod = 998244353;

int modexp(int a, int n)
{
    int b = 1;

    while (n)
    {
        if (n & 1)
        {
            b = 1LL * a * b % mod;
        }

        n >>= 1;
        a = 1LL * a * a % mod;
    }
    return b;
}

void add(int &a, int b)
{
    a += b;
    if (a >= mod)
        a -= mod;
}

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    int t18 = modexp(10, 18);

    int n;
    cin >> n;

    vector<int> p18(n + 1);
    p18[0] = 1;
    for (int i = 1; i <= n; ++i)
    {
        p18[i] = 1LL * p18[i - 1] * t18 % mod;
    }

    vector<int> grundy(1, 1);

    for (int i = 0; i < 3; ++i)
    {
        int m;
        cin >> m;

        vector<vector<int>> adj(n);

        for (int j = 0; j < m; ++j)
        {
            int u, v;
            cin >> u >> v;
            u--;
            v--;
            if (u > v)
                swap(u, v);
            adj[u].push_back(v);
        }

        vector<int> c_grundy(n);
        vector<int> n_grundy;

        int one = 0, zero = 0;

        for (int j = n - 1; j >= 0; --j)
        {
            set<int> s;
            for (auto u : adj[j])
                s.insert(c_grundy[u]);

            int g = 0;
            while (s.count(g))
                g++;

            c_grundy[j] = g;
            while (n_grundy.size() <= g)
                n_grundy.push_back(0);
            add(n_grundy[g], p18[j + 1]);
        }

        vector<int> a_grundy;

        for (int u = 0; u < grundy.size(); ++u)
        {
            for (int v = 0; v < n_grundy.size(); ++v)
            {
                int t = u ^ v;
                while (a_grundy.size() <= t)
                    a_grundy.push_back(0);

                add(a_grundy[t], 1LL * grundy[u] * n_grundy[v] % mod);
            }
        }

        grundy.swap(a_grundy);
    }

    cout << grundy[0] << endl;

    return 0;
}