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 <cstdio>
#include <vector>
#include <set>
#include <algorithm>

using namespace std;

const int mod = 998244353;
int m[3];
int g[3][100000];
int p[3];
long long sum[3][100000];
vector <int> v[3][100000];

long long powmod(long long x, long long y) {
    long long ans = 1;
    
    while (y > 0) {
        if (y & 1) ans = ans * x % mod;
        x = x * x % mod;
        y >>= 1;
    }
    
    return ans;
}

int main() {
    int n, i, j, k;
    long long ans = 0;
    
    scanf("%d", &n);
    
    for (i = 0; i < 3; i++) {
        scanf("%d", &m[i]);
        
        for (j = 0; j < m[i]; j++) {
            int x, y;
            
            scanf("%d %d", &x, &y);
            
            x--;
            y--;
            
            if (x > y) swap(x, y);
            
            v[i][x].push_back(y);
        }
    }
    
    for (i = 0; i < 3; i++) {
        for (j = n - 1; j >= 0; j--) {
            set <int> s;
            
            for (k = 0; k < v[i][j].size(); k++) s.insert(g[i][v[i][j][k]]);
            
            for (k = 0; ; k++) {
                if (!s.count(k)) {
                    g[i][j] = k;
                    break;
                }
            }
            
            p[i] = max(p[i], g[i][j]);
            sum[i][g[i][j]] += powmod(10, 18 * (j + 1));
            if (sum[i][g[i][j]] >= mod) sum[i][g[i][j]] -= mod;
        }
    }
    
    for (i = 0; i <= p[0]; i++) {
        for (j = 0; j <= p[1]; j++) {
            int x = i ^ j;
            
            if (x <= p[2]) ans += sum[0][i] * sum[1][j] % mod * sum[2][x] % mod;
        }
    }
    
    printf("%lld\n", ans % mod);
    
    return 0;
}
