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
import sys
from collections import defaultdict

n, *mabs = map(int, sys.stdin.buffer.read().split())
MOD = 998244353
base = 10 ** 18 % MOD
base_costs = [base]
for i in range(n - 1):
    base_costs.append(base_costs[-1] * base % MOD)
graphs = []
i = 0
for _ in range(3):
    m = mabs[i]
    i += 1
    j = i + 2 * m
    parents = [set() for _ in range(n)]
    for a, b in zip(mabs[i:j:2], mabs[i + 1:j:2]):
        a -= 1
        b -= 1
        if a > b:
            a, b = b, a
        parents[a].add(b)
    grundy = [0] * n
    weights = defaultdict(int)
    for v in range(n - 1, -1, -1):
        pgs = {grundy[p] for p in parents[v]}
        for g in range(m + 1):
            if g not in pgs:
                grundy[v] = g
                weights[g] += base_costs[v]
                break

    for g in weights:
        weights[g] %= MOD

    graphs.append(dict(weights))
    i = j

ans = 0
graph_x, graph_y, graph_z = graphs
for gx, cx in graph_x.items():
    for gy, cy in graph_y.items():
        gz = gx ^ gy
        if gz in graph_z:
            ans = (ans + cx * cy * graph_z[gz]) % MOD
print(ans)
