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 forn(i,n) for (int i = 0; i < int(n); i++)

const int MOD = 998244353;
const int N = 1e5 + 10;

int n, nim[3][N], cnt[3][N], mx[3], pot[N];
vector<int> adj[N];
bool vis[N];

void add(int &x, int y) {
    x += y;
    if (x >= MOD) x -= MOD;
}

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

    pot[0] = 1;
    pot[1] = 1; forn(_,18) pot[1] = 10LL*pot[1]%MOD;
    for (int i = 2; i < N; i++) pot[i] = (long long) pot[i-1] * pot[1] % MOD;

    forn(i,3) {
        forn(j,n) {
            adj[j].clear();
            nim[i][j] = 0;
        }
        int m; cin >> m;
        while (m--) {
            int u,v; cin >> u >> v;
            u--; v--;
            if (u > v) swap(u,v);
            adj[u].push_back(v);
        }
        for (int j = n-1; j >= 0; j--) {
            int sz = adj[j].size();
            fill_n(vis, sz+2, 0);
            for (auto k : adj[j]) vis[nim[i][k]] = 1;
            while (vis[nim[i][j]]) nim[i][j]++;

            add(cnt[i][nim[i][j]], pot[j+1]);
            mx[i] = max(mx[i], nim[i][j]);
        }
    }

    long long ans = 0;
    for (int x = 0; x <= mx[0]; x++) {
        for (int y = 0; y <= mx[1]; y++) {
            int z = x^y;
            ans += (long long) cnt[0][x] * cnt[1][y] % MOD * cnt[2][z] % MOD;
        }
    }
    cout << ans % MOD << endl;
    return 0;
}

