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;
#define int long long
#define ii pair <int, int>
#define app push_back
#define all(a) a.begin(), a.end()
#define bp __builtin_popcount
#define ll long long
#define mp make_pair
#define f first
#define s second
#define Time (double)clock()/CLOCKS_PER_SEC

const int MOD = 998244353;

int mod(int n) {
    n %= MOD;
    if (n < 0) return n + MOD;
    else return n;
}   
int fp(int a, int p) {
    int ans = 1, c = a;
    for (int i = 0; (1ll << i) <= p; ++i) {
        if ((p >> i) & 1) ans = mod(ans * c);
        c = mod(c * c);
    }   
    return ans;
}   
int dv(int a, int b) { return mod(a * fp(b, MOD - 2)); }

const int C = 1000, N = 1e5 + 7;
int sum[3][C];
int dp[N];
int n;

int b = fp(10, 18);
vector <int> g[N];
void solve(int t) {
    for (int i = 1; i <= n; ++i)
        g[i].clear();
    
    int m;
    cin >> m;
    for (int i = 0; i < m; ++i) {
        int u, v;
        cin >> u >> v;
        if (v < u)  
            swap(u, v);
        g[u].app(v);
    }   

    for (int i = n; i; --i) {
        vector <int> a;
        for (int v : g[i])
            a.app(dp[v]);
        sort(all(a));
        a.app(N);
        if (a[0])
            dp[i] = 0;
        else {
            for (int j = 0; ; ++j)
                if (a[j] + 1 < a[j + 1]) {
                    dp[i] = a[j] + 1;
                    break;
                }   
        }             
        sum[t][dp[i]] = mod(sum[t][dp[i]] + fp(b, i));
    }   

}   

signed main() {
    #ifdef HOME
    freopen("input.txt", "r", stdin);
    #else
    #define endl '\n'
    ios_base::sync_with_stdio(0); cin.tie(0);
    #endif
    cin >> n;
    for (int i = 0; i < 3; ++i)
        solve(i);
    int ans = 0;
    for (int a = 0; a < C; ++a)
        for (int b = 0; b < C; ++b) 
            ans = mod(ans + mod(mod(sum[0][a] * sum[1][b]) * sum[2][a ^ b]));
    cout << ans << endl;
}   