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>
#define mod 998244353
using namespace std;
int read();
int n;
int m1, m2, m3;
void Add(int& x, int y) { (x += y) >= mod ? x -= mod : x; }

vector<int> e[3][100005];
void add(int f, int t, int d) { e[d][f].push_back(t), e[d][t].push_back(f); }
int sg[3][100005];
vector<int> nx[100005];
void solve(int d) {
    for (int i = 1; i <= n; ++i) nx[i].clear();
    for (int u = n; u >= 1; --u) {
        sort(nx[u].begin(), nx[u].end());
        for (int v : nx[u]) {
            if (sg[d][u] == v) sg[d][u]++;
            if (sg[d][u] < v) break;
        }
        for (int v : e[d][u]) nx[v].push_back(sg[d][u]);
    }
}

int fsp(int base, int p) {
    int rt = 1;
    while (p) {
        if (p & 1) rt = 1ll * rt * base % mod;
        base = 1ll * base * base % mod, p >>= 1;
    }
    return rt;
}
int M(int x) { return x + (x >> 31 & mod); }
struct Poly {
    int x[300005];
    int& operator[](int p) { return x[p]; }
    void fwt(int len) {
        for (int l = 2; l <= len; l <<= 1)
            for (int i = 0, m = l >> 1; i < len; i += l)
                for (int j = i, t; j < i + m; ++j)
                    t = x[j + m], x[j + m] = M(x[j] - t), Add(x[j], t);
    }
} F[3];

int main() {
    n = read(), m1 = read();
    for (int i = 1; i <= m1; ++i) add(read(), read(), 0);
    m2 = read();
    for (int i = 1; i <= m2; ++i) add(read(), read(), 1);
    m3 = read();
    for (int i = 1; i <= m3; ++i) add(read(), read(), 2);
    for (int i = 0; i < 3; ++i) {
        solve(i);
        for (int j = 1, t = 1, p = fsp(10, 18); j <= n; ++j)
            t = 1ll * t * p % mod, Add(F[i][sg[i][j]], t);
    }
    int len = 1;
    while (len <= n) len <<= 1;
    F[0].fwt(len), F[1].fwt(len), F[2].fwt(len);
    for (int i = 0; i < len; ++i)
        F[0][i] = 1ll * F[0][i] * F[1][i] % mod * F[2][i] % mod;

    F[0].fwt(len);
    printf("%d\n", 1ll * F[0][0] * fsp(len, mod - 2) % mod);
    return 0;
}

int read() {
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') f = (c == '-') ? -1 : f, c = getchar();
    while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}