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
#include<bits/stdc++.h>

using namespace std;
typedef long long int lld;
#define rep(i,a,b) for(int i=a;i<b;i++)
#define trav(a,v) for(auto a:v)
#define MOD 998244353
vector<int> nei[1000000][3];
int used[1000000][3];

int main(){
  lld base=1;
  rep(i,0,18){
    base*=10;
    base%=MOD;
  }
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin>>n;
  rep(i,0,3){
    int m;
    cin>>m;
    rep(j,0,m){
      int x,y;
      cin>>x>>y;
      x--;y--;
      if(x>y)swap(x,y);
      nei[x][i].push_back(y);
    }
  }
  /*lld Res[3][2];
  rep(k,0,3){
    rep(i,0,2)Res[k][i]=0;
    rep(i,0,n)used[i][k]=-1;
    for(int i=n-1;i>-1;i--){
      if(used[i][k]==-1){
	used[i][k]=1;
	trav(a,nei[i][k]){
	  used[a][k]=0;
	}
      }
    }
    lld Pow=base;
    rep(i,0,n){
      if(used[i][k]){
	Res[k][1]+=Pow;
	Res[k][1]%=MOD;
      }else{
	Res[k][0]+=Pow;
	Res[k][0]%=MOD;
      }
      Pow*=base;
      Pow%=MOD;
    }
  }
  lld ans=0;
  rep(msk,0,8){
    lld can=1;
    int cnt=0;
    rep(i,0,3){
      if((msk>>i)%2>0)can*=Res[i][1],cnt++;
      else can*=Res[i][0];
      can%=MOD;
    }
    cnt%=2;
    //ans+=can;
    if(cnt==1){
      ans+=can;
      ans%=MOD;
    }
    }
    cout<<ans<<endl;*/
  int nim[n][3];
  int max_nim=0;
  set<int> s;
  rep(j,0,3){
    for(int i=n-1;i>-1;i--){
      s.clear();
      trav(a,nei[i][j]){
	s.insert(nim[a][j]);
      }
      int cnt=0;
      while(s.find(cnt)!=s.end())cnt++;
      nim[i][j]=cnt;
      max_nim=max(max_nim,cnt+1);
    }
  }
  lld Tot[max_nim][3];
  rep(i,0,max_nim){
    rep(j,0,3)Tot[i][j]=0;
  }
  //cout<<base<<endl;
  rep(j,0,3){
    lld Pow=base;
    rep(i,0,n){
      Tot[nim[i][j]][j]+=Pow;
      Tot[nim[i][j]][j]%=MOD;
      Pow*=base;
      Pow%=MOD;
    }
  }
  /*rep(i,0,max_nim){
    rep(j,0,3)cout<<Tot[i][j]<<" ";
    cout<<endl;
  }*/
  lld ans=0;
  rep(i,0,max_nim){
    rep(j,0,max_nim){
      if((i^j)<max_nim){
      lld can=1;
      can*=Tot[i][0];
      can%=MOD;
      can*=Tot[j][1];
      can%=MOD;
      can*=Tot[i^j][2];
      can%=MOD;
      ans+=can;
      ans%=MOD;
      //cout<<i<<" "<<j<<" "<<(i^j)<<endl;
      }
    }
  }
  cout<<ans<<endl;
  return 0;
}