The "BerCorp" company has got n employees. These employees can use m approved official languages for the formal correspondence. The languages are numbered with integers from 1 to m. For each employee we have the list of languages, which he knows. This list could be empty, i. e. an employee may know no official languages. But the employees are willing to learn any number of official languages, as long as the company pays their lessons. A study course in one language for one employee costs 1 berdollar.

Find the minimum sum of money the company needs to spend so as any employee could correspond to any other one (their correspondence can be indirect, i. e. other employees can help out translating).

Input

The first line contains two integers n and m (2 ≤ n, m ≤ 100) — the number of employees and the number of languages.

Then n lines follow — each employee's language list. At the beginning of the i-th line is integer ki (0 ≤ ki ≤ m) — the number of languages the i-th employee knows. Next, the i-th line contains ki integers — aij (1 ≤ aij ≤ m) — the identifiers of languages the i-th employee knows. It is guaranteed that all the identifiers in one list are distinct. Note that an employee may know zero languages.

The numbers in the lines are separated by single spaces.

Output

Print a single integer — the minimum amount of money to pay so that in the end every employee could write a letter to every other one (other employees can help out translating).

Examples

Input

5 5
1 2
2 2 3
2 3 4
2 4 5
1 5


Output

0


Input

8 7
0
3 1 2 3
1 1
2 5 4
2 6 7
1 3
2 7 4
1 1


Output

2


Input

2 2
1 2
0


Output

1

Note

In the second sample the employee 1 can learn language 2, and employee 8 can learn language 4.

In the third sample employee 2 must learn language 2.
#######################################################################################################################
#   Author:     BlackFyre
#   Language:   PyPy 3.7
#######################################################################################################################
 
from sys import stdin, stdout, setrecursionlimit
from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log, log2
from random import seed, randint
from datetime import datetime
from collections import defaultdict as dd, deque
from heapq import merge, heapify, heappop, heappush, nsmallest
from bisect import bisect_left as bl, bisect_right as br, bisect
from collections import defaultdict as dd
mod = pow(10, 9) + 7
mod2 = 998244353
# setrecursionlimit(3000)

def inp(): return stdin.readline().strip()
def iinp(): return int(inp())
def out(var, end="\n"): stdout.write(str(var) + "\n")
def outa(*var, end="\n"): stdout.write(' '.join(map(str, var)) + end)
def lmp(): return list(mp())
def mp(): return map(int, inp().split())
def smp(): return map(str, inp().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(m, val) for j in range(n)]
def remadd(x, y): return 1 if x % y else 0
def ceil(a, b): return (a + b - 1) // b
def def_value(): return 0
def def_inf(): return inf

n,m = mp()
adj = dd(list)
for i in range(n):
    adj[i+1].append(i)
    adj[i+1].pop()
mc = []
for i in range(n):
    a = lmp()
    mc.append(a[0])
    for j in a[1:]:
        adj[i+1].append(j+n)
        adj[j+n].append(i+1)

if sum(mc)>0:
    ans = -1
    mark = l1d(n+m+1)
    st = []
    for i in adj.keys():
        if mark[i]==0:
            ans+=1
            mark[i]+=1
            st.append(i)
            while len(st)>0:
                x = st.pop()
                for y in adj[x]:
                    if mark[y]==0:
                        mark[y]+=1
                        st.append(y)
    print(ans)
else: print(n)