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 <iostream>
#include <stdio.h>
#include <string.h>
#include <map>
#include <unordered_map>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <math.h>
#include <time.h>
#include <assert.h>

using namespace std;

#define x first
#define y second
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define pii pair<int, int>
#define all(x) x.begin(),x.end()
#define SZ(x) int(x.size())
#define rep(i, a, b) for(int i = a; i < b; i++)
#define per(i, a, b) for(int i = b-1; i >= a; i--)
#define DBG(x) cerr << (#x) << "=" << x << "\n";
#define ll long long
#define inf 1000000007
#define mod 998244353//1000000007
#define N 100005

template<typename U, typename V> void Min(U &a, const V & b){if(a > b)a = b;}
template<typename U, typename V> void Max(U &a, const V & b){if(a < b)a = b;}
template<typename U, typename V> void add(U &a, const V & b){a = (a + b) % mod;}
template<typename U> U gcd(U a, U b){
    if(a == 0)return b;
    if(b == 0)return a;
    if(a >= b)return gcd(a % b, b);
    else return gcd(a, b % a);
}
int pow(int a, int b){
    int ans = 1;
    while(b){
        if(b & 1)ans = 1LL * ans * a % mod;
        a = 1LL * a * a % mod;
        b >>= 1;
    }
    return ans;
}
int pow(int a, ll b, int c){
    int ans = 1;
    while(b){
        if(b & 1)ans = 1LL * ans * a % c;
        a = 1LL * a * a % c;
        b >>= 1;
    }
    return ans;
}

int vis[N], sg[N], dp[3][N], p[N], now, mx[3];
vi g[N];
int main(){
  int T, m, i, j, k, n, K;
  scanf("%d", &n);
  p[0] = 1;
  p[1] = pow(10, 18);
  rep(i, 2, n+1)p[i] = 1LL * p[i-1] * p[1] % mod;
  rep(it, 0, 3){
    scanf("%d", &m);
    rep(i, 0, n)g[i].clear();
    while(m--){
      scanf("%d%d", &i, &j);
      i--, j--;
      if(i > j)swap(i, j);
      g[i].pb(j);
    }
    per(i, 0, n){
      now++;
      sg[i] = 0;
      for(auto &v : g[i])vis[sg[v]] = now;
      while(vis[sg[i]] == now)sg[i]++;
      Max(mx[it], sg[i]);
      add(dp[it][sg[i]], p[i+1]);
    }
  }
  ll ans = 0;
  rep(i, 0, mx[0]+1){
    rep(j, 0, mx[1]+1){
      ans = ans + 1LL * dp[0][i] * dp[1][j] % mod * dp[2][i^j] % mod;
    }
  }
  ans %= mod;
  printf("%lld\n", ans);
}
