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;

const int M = 998244353;
const int N = 1e5+7;
const long long B = 1000000000000000000;
typedef long long LL;

LL pb[N];
void pre() {
    pb[0] = 1;
    LL b = B%M;
    for (int i=1; i<N; i++)
        pb[i] = (pb[i-1]*b)%M;
}

vector<int> bfs(vector<vector<int>>adj) {
    int n = adj.size()-1;
    vector<int> col(n+1, N);

    vector<bool> vis(n);
    for (int i=n; i>0; i--) {
        for (int v: adj[i]) {
            if (v > i)    vis[col[v]] = 1;
        }

        for (int ii=0; ii<=n; ii++)
            if (!vis[ii]) {
                col[i] = ii;
                break;
            }

        for (int v: adj[i]) {
            if (v > i)    vis[col[v]] = 0;
        }
    }
    return col;
}

int m[3];
vector<vector<int>> adj[3];
vector<int> dis[3];
vector<LL> con[3];

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    pre();

    int n;
    cin>>n;
    const int MX = 1<<17;

    for (int g=0; g<3; g++) {
        cin>>m[g];
        adj[g].resize(n+1);
        for (int i=0; i<m[g]; i++) {
            int u, v;
            cin>>u>>v;
            adj[g][u].push_back(v);
            adj[g][v].push_back(u);
        }
        dis[g] = bfs(adj[g]);

        con[g].resize(MX);
        for (int i=1; i<=n; i++) {
            con[g][dis[g][i]] = (con[g][dis[g][i]] + pb[i])%M;
        }
    }
    LL ans = 0;
    for (int i=0; i<MX; i++) {
        if (con[0][i] == 0) continue;
        for (int j=0; j<MX; j++) {
            if (con[1][j] == 0)  continue;
            ans = (ans + ((con[0][i]*con[1][j])%M)*con[2][i^j])%M;
        }
    }
    cout<<ans<<endl;
}