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;
#define mod 998244353
long long p[100010];

void init() {
    p[0] = p[1] = 1;
    for (int i = 0; i < 18; i++) {
        p[1] = p[1] * 10 % mod;
    }
    for (int i = 1; i <= 100000; i++) {
        p[i+1] = p[i] * p[1] % mod;
    }
}

void calc(int n, vector<long long> &v) {
    int m; cin >> m;
    vector<int> e[n+1];
    vector<int> g(n+1, -1);
    for (int i = 0; i < m; i++) {
        int a, b; cin >> a >> b;
        e[a].emplace_back(b);
        e[b].emplace_back(a);
    }
    for (int i = n; i > 0; i--) {
        set<int> s;
        for (auto &to: e[i]) {
            s.emplace(g[to]);
        }
        int t = 0;
        for (auto &x: s) {
            if (x == -1) continue;
            if (x != t) break;
            t++;
        }
        g[i] = t;
    }
    for (int i = 1; i <= n; i++) {
        v[g[i]] = (v[g[i]] + p[i]) % mod;
    }
}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    init();
    int n; cin >> n;
    vector<long long> x(512, 0), y(512, 0), z(512, 0);
    calc(n, x); calc(n, y); calc(n, z);
    long long ans = 0;
    for (int i = 0; i < 512; i++) {
        for (int j = 0; j < 512; j++) {
            ans = (ans + x[i] * y[j] % mod * z[i^j] % mod) % mod;
        }
    }
    cout << ans << "\n";
    return 0;
}