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<iostream>
#include<cstdio>
#include<algorithm>
#include<cassert>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<stack>
using namespace std;
#define MOD 1000000007
#define MOD2 998244353
#define INF ((1<<30)-1)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long Int;
typedef pair<Int, Int> P;

vector<Int> edge[3][110000];
Int grundy[3][110000];
Int grundysum[3][110000];
Int n, m[3], a, b;
bool nums[110000];

Int mod_pow(Int x, Int a, Int m = MOD){
    if(a == 0)return 1;
    Int res = mod_pow(x, a / 2, m);
    res = res * res % m;
    if(a % 2)res *= x;
    return res % m;
}

Int inv(Int x, Int m = MOD){
    return mod_pow(x, m-2, m);
}


void calc(Int x){
    for(Int i = n;i >= 1;i--){
        for(Int neib:edge[x][i]){
            if(neib < i)continue;
            nums[grundy[x][neib]] = true;
        }
        while(nums[grundy[x][i]])grundy[x][i]++;
        for(Int neib:edge[x][i]){
            if(neib < i)continue;
            nums[grundy[x][neib]] = false;
        }
        grundysum[x][grundy[x][i]] += mod_pow(10, 18 * i, MOD2);
        grundysum[x][grundy[x][i]] %= MOD2;
    }
}

int main(){
    cin >> n;
    for(Int i = 0;i < 3;i++){
        cin >> m[i];
        for(Int j = 0;j < m[i];j++){
            cin >> a >> b;
            edge[i][a].push_back(b);
            edge[i][b].push_back(a);
        }
        calc(i);
    }

    Int ans = 0;
    for(Int i = 0;i < 600;i++){
        for(Int j = 0;j < 600;j++){
            ans += grundysum[0][i] * grundysum[1][j] % MOD2 * grundysum[2][i^j] % MOD2;
            ans %= MOD2;
        }
    }
    cout << ans << endl;
    return 0;
}