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
from collections import defaultdict

M = 998244353
B = 10**18 % M

def mex(s):
  for i in range(max(s)+2):
    if i not in s:
      return i

def ext_euc(a, b):
  x1, y1, z1 = 1, 0, a
  x2, y2, z2 = 0, 1, b
  while z1 != 1:
    d, m = divmod(z2,z1)
    x1, x2 = x2-d*x1, x1
    y1, y2 = y2-d*y1, y1
    z1, z2 = m, z1
  return x1, y1

def inv_mod(a, b, m):
  x, y = ext_euc(a, m)
  return (x * b % m)

def calc_grundy(e):
  g = {}
  sum_g = defaultdict(int)
  sum_g[0] = inv_mod(B-1, pow(B, N+1, M)-B, M)
  for i in sorted(e.keys(), reverse=True):
    m = mex({g.get(j, 0) for j in e[i]})
    if m:
      g[i] = m
      x = pow(B, i, M)
      sum_g[m] += x
      sum_g[0] -= x
  return sum_g

def read_edge():
  M = int(input())
  e = defaultdict(set)
  for i in range(M):
    a, b = sorted(map(int, input().split()))
    e[a].add(b)
  return e

def solve(N, edge):
  sum_g = list(map(calc_grundy, edge))
  ret = 0
  for gx, sx in sum_g[0].items():
    for gy, sy in sum_g[1].items():
      gz = gx^gy
      sz = sum_g[2][gz]
      if sz:
        ret = (ret + sx*sy*sz)%M
  return ret

N = int(input())
edge = [read_edge() for i in range(3)]

print(solve(N, edge))