import sys
input = lambda : sys.stdin.readline().rstrip()

sys.setrecursionlimit(2*10**5+10)
write = lambda x: sys.stdout.write(x+"\n")
debug = lambda x: sys.stderr.write(x+"\n")
writef = lambda x: print("{:.12f}".format(x))


n = int(input())
hw = []
m = 0
for _ in range(n):
    h,w = list(map(int, input().split()))
    hw.append((h,w))
    m = max(m, h, w)
ns = [[] for _ in range(m)]
rns = [[] for _ in range(m)]
u = 0
for h,w in hw:
    ns[h-1].append(w-1)
    rns[w-1].append(h-1)
#     ns[u].append(n+w-1)
#     ns[n+h-1].append(u)
#     rns[n+w-1].append(u)
#     rns[u].append(n+h-1)
    u += 1
v0 = v1 = 0
start = None
end = None
for u in range(m):
    if len(ns[u])==len(rns[u]):
        pass
    elif len(ns[u])+1 == len(rns[u]):
        v0 += 1
        end = u
    elif len(rns[u])+1 == len(ns[u]):
        v1 += 1
        start = u
    else:
        v0 = v1 = 100
        break
def euler_d(ns, start=0):
    """有向グラフのオイラー閉路
    パスの場合はstartを奇数次数頂点にする
    """
    n = len(ns)
    p = []
    q = [start]
    r = []
    vs = [0]*n
    while q:
        u = q.pop()
        u0 = u
        while vs[u]<len(ns[u]):
            v = ns[u][vs[u]]
            vs[u] += 1
            u = v
            r.append(v)
        while r:
            q.append(r.pop())
        p.append(u0)
    return p
if v0==v1==0:
    # ひとつかどうか判定
    for start in range(len(ns)):
        if ns[start]:
            break
    res = euler_d(ns, start)
    if len(res)==n+1:
        s = set()
        l = res[:-1]
        for i in range(len(l)):
            h0 = l[i]
            w1 = l[i-1]
            s.add((h0))
        ans = len(s)
    else:
        ans = 0
elif v0==v1==1:
    ns[end].append(start)
    res = euler_d(ns, start)
    if len(res)==n+2:
        ans = 1
    else:
        ans = 0
else:
    ans = 0
print(ans)