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>

#define long long long int
static const long MOD = 998244353;
using namespace std;

// @author: pashka

int main() {
    ios::sync_with_stdio(false);

    int n;
    cin >> n;
    vector<vector<vector<int>>> g(3, vector<vector<int>>(n));
    for (int t = 0; t < 3; t++) {
        int m;
        cin >> m;
        for (int i = 0; i < m; i++) {
            int x, y;
            cin >> x >> y;
            x--;
            y--;
            if (x > y) swap(x, y);
            g[t][x].push_back(y);
        }
    }
    vector<vector<int>> gr(3, vector<int>(n));
    int mx = 0;
    for (int t = 0; t < 3; t++) {
        vector<int> z(n + 1);
        int zz = 0;
        for (int i = n - 1; i >= 0; i--) {
            zz++;
            for (int j : g[t][i]) {
                z[gr[t][j]] = zz;
            }
            while (z[gr[t][i]] == zz) gr[t][i]++;
            mx = max(mx, gr[t][i]);
        }
    }
    vector<vector<long>> ss(3, vector<long>(mx + 1));
    vector<long> pw(n + 1);
    pw[0] = 1;
    for (int i = 0; i < n; i++) {
        pw[i + 1] = pw[i] * 716070898 % MOD;
    }
    for (int t = 0; t < 3; t++) {
        for (int i = 0; i < n; i++) {
            ss[t][gr[t][i]] += pw[i + 1];
            ss[t][gr[t][i]] %= MOD;
        }
    }
    long res = 0;
    for (int i = 0; i <= mx; i++) {
        for (int j = 0; j <= mx; j++) {
            if ((i ^ j) > mx) continue;
            long s = ss[0][i];
            s *= ss[1][j];
            s %= MOD;
            s *= ss[2][i ^ j];
            s %= MOD;
            res += s;
            res %= MOD;
        }
    }
    cout << res << "\n";
    return 0;
}