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;

const int maxn = 1e5 + 3;
const int mod = 998244353;

int n, m[4];
vector <int> adj[4][maxn];
void read()
{
    cin >> n;
    for (int i = 1; i <= 3; i++)
    {
        cin >> m[i];
        for (int j = 1; j <= m[i]; j++)
        {
            int a, b;
            cin >> a >> b;

            adj[i][a].push_back(b);
            adj[i][b].push_back(a);
        }
    }
}

long long add(long long a, long long b)
{
    a += b;
    if (a >= mod)
        a -= mod;

    return a;
}

long long st[maxn*2];
int gr[4][maxn];
map <int, long long> mp[4];
void solve()
{
    st[0] = 1;
    for (int i = 1; i <= 2 * n; i++)
        st[i] = st[i-1] * 1000000000 % mod;

    for (int i = 1; i <= 3; i++)
    {
        mp[i][0] = add(mp[i][0], st[n*2]);
        for (int j = n-1; j >= 1; j--)
        {
            set <int> s;
            for (auto k: adj[i][j])
                if (k > j)
                s.insert(gr[i][k]);

            while (s.find(gr[i][j]) != s.end())
                gr[i][j]++;

            mp[i][gr[i][j]] = add(mp[i][gr[i][j]], st[j*2]);
        }
    }

    int p = sqrt(max({m[1], m[2], m[3]})) * 2;
    long long ans = 0;
    for (int i = 0; i <= p; i++)
        if (mp[1].count(i))
        for (int j = 0; j <= p; j++)
        if (mp[2].count(j))
            ans = add(ans, mp[1][i] * mp[2][j] % mod * mp[3][i^j] % mod);

    cout << ans << endl;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    read();
    solve();
}
