from math import sqrt,sin,cos,tan,ceil,radians,floor,gcd,exp,log,log10,log2,factorial,fsum
from heapq import heapify, heappop, heappush
from bisect import bisect_left, bisect_right
from copy import deepcopy
import copy
import random
from collections import deque,Counter,defaultdict
from itertools import permutations,combinations
from decimal import Decimal,ROUND_HALF_UP
#tmp = Decimal(mid).quantize(Decimal('0'), rounding=ROUND_HALF_UP)
from functools import lru_cache, reduce
#@lru_cache(maxsize=None)
from operator import add,sub,mul,xor,and_,or_,itemgetter
INF = 10**18
mod1 = 10**9+7
mod2 = 998244353

#DecimalならPython
#再帰ならPython!!!!!!!!!!!!!!!!!!!!!!!!!!

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n
        #親だけ
        self.edge = [0]*n

    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]

    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if self.parents[x] > self.parents[y]:
            x, y = y, x
        
        self.edge[x] += 1
        
        if x == y:
            return

        self.parents[x] += self.parents[y]
        self.edge[x] += self.edge[y]
        self.parents[y] = x

    def size(self, x):
        return -self.parents[self.find(x)]

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):
        return len(self.roots())

    def all_group_members(self):
        group_members = defaultdict(list)
        for member in range(self.n):
            group_members[self.find(member)].append(member)
        return group_members

    def __str__(self):
        return '\n'.join(f'{r}: {m}' for r, m in self.all_group_members().items())

'''
グラフっぽく考えられそう
最初と終わりさえ決めれば間は関係ない
全部を通って終わりまでたどりつけるか？という問題

sccして1個なら全部行ける
1個じゃなくて、パスがあったらその組はいける

入次数0から始める
そこから全部通れるか
サイクルはSCCで対処可能
'''

N = int(input())
lim = 2*(10**5)+1
lef = [[] for _ in range(lim)]
rig = [[] for _ in range(lim)]

mem = set()
hw = []
for i in range(N):
    h,w = map(int, input().split())
    hw.append((h,w))
    lef[h].append((w,i))
    mem.add(h)
    mem.add(w)

M = len(mem)
uf = UnionFind(N)

memo1 = [0]*(lim)
memo2 = [0]*(lim)
i = 0
edge = []
for h,w in hw:
    for l,j in lef[w]:
        memo1[j] += 1
        memo2[i] += 1
        uf.union(i,j)
        edge.append((h,l))
    i += 1

if uf.group_count() != 1:
    print(0)
    exit()

ls = []
for i in range(N):
    ls.append(memo1[i]-memo2[i])

ls.sort()

if ls[0] == ls[-1] == 0:
    print(M)
elif ls[0] == -ls[-1] and ls[1] == ls[-2] == 0:
    print(1)
else:
    print(0)