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;
using LL = long long;
constexpr LL mod = 998244353;
constexpr int maxn = 100000;
int N, sg[maxn + 1], s[maxn + 1];
LL p[maxn + 1];
vector<int> G[maxn + 1];
vector<LL> g(){
    vector<LL> res;
    int M;
    cin >> M;
    for(int i = 1; i <= N; i += 1) G[i].clear();
    for(int i = 0, u, v; i < M; i += 1){
        cin >> u >> v;
        if(u > v) swap(u, v);
        G[u].push_back(v);
    }
    for(int i = N; i; i -= 1){
        sg[i] = 0;
        for(int v : G[i]) s[sg[v]] = 1;
        while(s[sg[i]]) sg[i] += 1;
        for(int v : G[i]) s[sg[v]] = 0;
        if(res.size() <= sg[i]) res.resize(sg[i] + 1);
        res[sg[i]] = (res[sg[i]] + p[i]) % mod;
    }
    return res;
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    cin >> N;
    p[1] = 1000000000000000000LL % mod;
    for(int i = 2; i <= N; i += 1) p[i] = p[i - 1] * p[1] % mod;
    auto g1 = g();
    auto g2 = g();
    auto g3 = g();
    LL ans = 0;
    for(int i = 0; i < (int)g1.size(); i += 1)
        for(int j = 0; j < (int)g2.size(); j += 1){
            int k = i ^ j;
            if(k < (int)g3.size()) ans += g1[i] * g2[j] % mod * g3[k] % mod;
        }
    cout << ans % mod;
    return 0;
}