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;
typedef complex<long double> base;
const double pi=acos(-1.);
const int MX=132132,md=998244353;
int n,m,it,ut,i,x,y,res,u[MX],a[3][MX],mx[3];
long long cnt[3][MX];
vector<int> g[MX];
long long pw10(long long x) {
  if (x==0) return 1LL;
  if (x&1) return (pw10(x-1)*10LL)%md;
  long long z=pw10(x/2);
  return (z*z)%md;
}
int main() {
  scanf("%d",&n);
  for (it=0; it<3; it++) {
    for (i=1; i<=n; i++) g[i].clear();
    scanf("%d",&m);
    while (m--) {
      scanf("%d%d",&x,&y);
      g[min(x,y)].push_back(max(x,y));
    }
    for (i=n; i>0; i--) {
      ++ut;
      for (x=0; x<g[i].size(); x++) u[a[it][g[i][x]]]=ut;
      for (a[it][i]=0; u[a[it][i]]==ut; a[it][i]++);
      cnt[it][a[it][i]]=(cnt[it][a[it][i]]+pw10(18*i))%md;
      mx[it]=max(mx[it],a[it][i]);
    }
  }
  for (i=0; i<=mx[0]; i++) if (cnt[0][i]) for (x=0; x<=mx[1]; x++) if (cnt[1][x] && cnt[2][i^x])
    res=(res+((cnt[0][i]*cnt[1][x])%md)*cnt[2][i^x])%md;
  printf("%d\n",res);
  return 0;
}