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>

#define ll long long
#define fornum(A,B,C) for(A=B;A<C;++A)
#define pii pair<int,int>
#define pll pair<ll,ll>

using namespace std;

/////////////////////////////////////////////////////
#define MOD 998244353
 


ll N,M[5];
vector<ll> vv[5][101010];
ll gr[5][101010];
vector<ll> grmk[5][101010];
ll mk[101010], mk2[101010],mk3[101010];
ll pans[505050];
ll i, j, k;
ll ans;


int main(){
    scanf("%lld", &N);
    fornum(i,0,3){
        scanf("%lld", &M[i]);
        fornum(j,0,M[i]){
            ll a, b;
            scanf("%lld%lld", &a, &b);
            vv[i][a].push_back(b);
            vv[i][b].push_back(a);
        }
    }
    pans[0] = 1;
    pans[1] = (ll)1e18 % MOD;
    fornum(i,0,3*N){
        pans[i + 1] = pans[i] * pans[1] % MOD;
    }

    fornum(i,0,3){
        for (j = N; j >= 1;--j){
            ll a = 0;
            sort(grmk[i][j].begin(), grmk[i][j].end());
            fornum(k,0,grmk[i][j].size()){
                if(a<grmk[i][j][k]){
                    break;
                }
                a = grmk[i][j][k] + 1;
            }
            gr[i][j] = a;
            fornum(k,0,vv[i][j].size()){
                ll b = vv[i][j][k];
                grmk[i][b].push_back(gr[i][j]);
            }
        }
    }
    mk[0] = 1;
    fornum(i,0,3){
        fornum(j,0,1010){
            mk2[j] = mk[j];
            mk3[j] = mk[j] = 0;
        }
        fornum(j,0,N){
            mk3[gr[i][j + 1]] += pans[j + 1];
            mk3[gr[i][j + 1]] %= MOD;
        }
        fornum(j,0,1010){
            fornum(k,0,1010){
                mk[j ^ k] += mk2[j] * mk3[k] % MOD;
                mk[j ^ k] %= MOD;
            }
        }
        
    }
    printf("%lld\n", mk[0]);
    return 0;
}