Problem: 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

Solution (in PYTHON3):

Solution:
import itertools
import os
import sys
from collections import defaultdict

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(10 ** 9)
INF = float("inf")
IINF = 10 ** 18
# MOD = 10 ** 9 + 7
MOD = 998244353

# 解説
# 大きい方から貪欲に取るのの見方を変えるとゲームになる

N = int(sys.stdin.buffer.readline())
M = []
m = int(sys.stdin.buffer.readline())
M.append(m)
XVU = [list(map(int, sys.stdin.buffer.readline().split())) for _ in range(m)]
m = int(sys.stdin.buffer.readline())
M.append(m)
YVU = [list(map(int, sys.stdin.buffer.readline().split())) for _ in range(m)]
m = int(sys.stdin.buffer.readline())
M.append(m)
ZVU = [list(map(int, sys.stdin.buffer.readline().split())) for _ in range(m)]

X = [[] for _ in range(N)]
Y = [[] for _ in range(N)]
Z = [[] for _ in range(N)]
for G, VU in zip([X, Y, Z], (XVU, YVU, ZVU)):
    for v, u in VU:
        v -= 1
        u -= 1
        if v > u:
            v, u = u, v
        G[v].append(u)


def grundy(graph):
    ret = [-1] * N
    for v in reversed(range(N)):
        gs = set()
        for u in graph[v]:
            gs.add(ret[u])
        g = 0
        while g in gs:
            g += 1
        ret[v] = g
    return ret


grundy_x = defaultdict(list)
grundy_y = defaultdict(list)
grundy_z = defaultdict(list)
for x, g in enumerate(grundy(X), 1):
    grundy_x[g].append(x)
for y, g in enumerate(grundy(Y), 1):
    grundy_y[g].append(y)
for z, g in enumerate(grundy(Z), 1):
    grundy_z[g].append(z)

# base[i]: (10^18)^i
base = [1]
for _ in range(N * 3 + 10):
    base.append(base[-1] * pow(10, 18, MOD) % MOD)

ans = 0
for xg, yg in itertools.product(grundy_x.keys(), grundy_y.keys()):
    zg = xg ^ yg
    # 10^(x1+y1+z1) + 10^(x1+y1+z2) + ... + 10^(x1+y2+z1) + ...
    # = (10^x1 + 10^x2 + ...) * (10^y1 + ...) * (10^z1 + ...)
    xs = 0
    ys = 0
    zs = 0
    for x in grundy_x[xg]:
        xs += base[x]
    for y in grundy_y[yg]:
        ys += base[y]
    for z in grundy_z[zg]:
        zs += base[z]
    ans += xs % MOD * ys % MOD * zs % MOD
    ans %= MOD
print(ans)