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;

typedef long long ll;
typedef pair <int, int> ii;

const int Maxn = 100005;
const int mod = 998244353;
const int pw = 1000000;
const int Maxm = 1005;

int n;
vector <int> neigh[Maxn];
vector <int> my[3];
int col[3][Maxn];
int res;
bool was[3][Maxm][Maxm];

int toPower(int a, int p)
{
    int res = 1;
    while (p) {
        if (p & 1) res = ll(res) * a % mod;
        p >>= 1; a = ll(a) * a % mod;
    }
    return res;
}

void Solve(vector <int> &res, int col[])
{
    int m; scanf("%d", &m);
    for (int i = 1; i <= n; i++)
        neigh[i].clear();
    for (int i = 0; i < m; i++) {
        int a, b; scanf("%d %d", &a, &b);
        if (a > b) swap(a, b);
        neigh[a].push_back(b);
    }
    for (int i = n; i > 0; i--) {
        vector <int> seq;
        for (int j = 0; j < neigh[i].size(); j++)
            seq.push_back(col[neigh[i][j]]);
        sort(seq.begin(), seq.end());
        col[i] = 0;
        int my = toPower(pw, 3 * i);
        for (int j = 0; j < seq.size(); j++)
            if (seq[j] == col[i]) col[i]++;
        if (col[i] == res.size()) res.push_back(my);
        else res[col[i]] = (res[col[i]] + my) % mod;
    }
}

int main()
{
    scanf("%d", &n);
    for (int i = 0; i < 3; i++) {
        Solve(my[i], col[i]);
        assert(my[i].size() < Maxm);
    }
    for (int i = 0; i < my[0].size(); i++)
        for (int j = 0; j < my[1].size(); j++)
            for (int k = 0; k < my[2].size(); k++) {
                if (was[0][i][j] || was[1][i][k] || was[2][j][k]) continue;
                was[0][i][j] = was[1][i][k] = was[2][j][k] = true;
                res = (res + ll(my[0][i]) * my[1][j] % mod * my[2][k]) % mod;
            }
    printf("%d\n", res);
    return 0;
}