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:
// No.074901032320
#include <bits/stdc++.h>
#define N 100100
using namespace std;
const int Mod = 998244353;

int n, m, vis[N], sg[N], f[3][N], mx, pw[N];
vector<int> G[N];

int main() {
#ifdef isLOCAL
    freopen(".in", "r", stdin);
    freopen(".out", "w", stdout);
#endif

    cin >> n;

    int base = 1000000000000000000ll % Mod;
    for (int i = pw[0] = 1; i <= n; ++i)
        pw[i] = 1ll * pw[i - 1] * base % Mod;

    for (int g = 0; g < 3; ++g) {
        cin >> m;
        for (int i = 1; i <= n; ++i) G[i].clear();
        for (int i = 1, u, v; i <= m; ++i) {
            scanf("%d%d", &u, &v);
            if (u > v) swap(u, v);
            G[u].push_back(v);
        }
        for (int i = n; i >= 1; --i) {
            for (int v : G[i])
                vis[sg[v]]++;
            sg[i] = 0;
            while (vis[sg[i]]) ++sg[i];
            for (int v : G[i])
                vis[sg[v]]--;
            mx = max(mx, sg[i]);
            f[g][sg[i]] = (f[g][sg[i]] + pw[i]) % Mod;
        }
    }
    
    long long ans = 0;
    for (int i = 0; i <= mx; ++i)
        for (int j = 0; j <= mx; ++j)
            ans = (ans + 1ll * f[0][i] * f[1][j] % Mod * f[2][i xor j]) % Mod;
    cout << ans << endl;

    return 0;
}