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 ll long long

ll mod = 998244353;

int main() {
    int N;
    cin >> N;
    
    vector<int> M(3);
    vector<vector<vector<int>>> edge(3, vector<vector<int>>(N+1));
    for (int i = 0; i < 3; i++) {
        cin >> M[i];
        for (int j = 0; j < M[i]; j++) {
            int a, b;
            cin >> a >> b;
            edge[i][b].push_back(a);
            edge[i][a].push_back(b);
        }
    }
    
    ll q = 1e18;
    q %= mod;
    vector<ll> pow10(N+1, 1);
    for (int i = 1; i <= N; i++) {
        pow10[i] = pow10[i-1] * q % mod;
    }
    
    vector<vector<int>> grundy(3, vector<int>(N+1, 0));
    vector<vector<ll>> gsum(3, vector<ll>(512, 0));
    for (int i = 0; i < 3; i++) {
        for (int j = N; j >= 1; j--) {
            vector<bool> check(N, false);
            for (int nxt : edge[i][j]) {
                if (nxt < j) continue;
                check[grundy[i][nxt]] = true;
            }
            int pos = 0;
            while (check[pos]) pos++;
            grundy[i][j] = pos;
            gsum[i][pos] = (gsum[i][pos] + pow10[j])%mod;
        }
    }
    
    ll ans = 0;
    for (int i = 0; i < 512; i++) {
        for (int j = 0; j < 512; j++) {
            ans += gsum[0][i]*gsum[1][j]%mod*gsum[2][i^j]%mod;
            ans %= mod;
        }
    }
    cout << ans << endl;
    return 0;
}