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 <cstdio>
#include <cctype>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
#define LOG(...) fprintf(stderr, __VA_ARGS__)
using namespace std;
typedef long long i64;
inline int rd(int f = 1, int x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 1e5+5, P = 998244353, W = int(((long long)1e18)%P);
int n, ans; void fix(int &x) { x<P?0:x-=P; }
struct G
{
    int sg[N], w[N], t; vector<int> e[N]; void add(int x, int y) { if(x > y) swap(x, y); e[x].push_back(y); }
    int SG(int x)
    {
        if(~sg[x]) return sg[x]; sg[x] = 0; map<int, bool> h;
        for(int y:e[x]) h[SG(y)] = 1; while(h.count(sg[x])) ++sg[x]; return sg[x];
    }
    void sol()
    {
        memset(sg+1, -1, sizeof(int)*n);
        for(int i = 1; i <= n; ++i) sg[i] = SG(i);
        for(int i = 1, v = 1; i <= n; ++i) v = (i64)v*W%P, t = max(sg[i], t), fix(w[sg[i]]+=v);
    }
}g[3];
int main()
{
    n = rd(); for(int k = 0; k < 3; g[k].sol(), ++k) for(int m = rd(), x, y; m--; ) x = rd(), y = rd(), g[k].add(x, y);
    for(int i = 0; i <= g[0].t; ++i) for(int j = 0; j <= g[1].t; ++j) fix(ans+=(i64)g[0].w[i]*g[1].w[j]%P*g[2].w[i^j]%P);
    printf("%d\n", ans);
    return 0;
}