# Original solution
class UnionFindNode(object):
    """
    Union-Find構造
    ノードのグループ併合や、所属グループ判定を高速に処理する
    """

    def __init__(self, group_id, parent=None, value=None):
        self.group_id_ = group_id
        self.parent_ = parent
        self.value = value
        self.size_ = 1

    def __str__(self):
        template = "UnionFindNode(group_id: {}, \n\tparent: {}, value: {}, size: {})"
        return template.format(self.group_id_, self.parent_, self.value, self.size_)

    def is_root(self):
        return not self.parent_

    def root(self):
        parent = self
        while not parent.is_root():
            parent = parent.parent_
            self.parent_ = parent
        return parent

    def find(self):
        parent = self.root()
        return parent.group_id_

    def size(self):
        parent = self.root()
        return parent.size_

    def unite(self, unite_node):
        parent = self.root()
        unite_parent = unite_node.root()

        if parent.group_id_ != unite_parent.group_id_:
            if parent.size() > unite_parent.size():
                unite_parent.parent_ = parent
                parent.size_ = parent.size_ + unite_parent.size_
            else:
                parent.parent_ = unite_parent
                unite_parent.size_ = parent.size_ + unite_parent.size_
    def is_group(self,other):#selfとotherが同じグループに属するか
        return self.root()==other.root()


N=int(eval(input()))
X=[]
Y=[]
for i in range(N):
    x,y=list(map(int,input().split()))
    X.append((x,i))
    Y.append((y,i))
X.sort()
Y.sort()
E=[]
ans=0
count=1
for i in range(N-1):
    a=X[i]
    b=X[i+1]
    c=Y[i]
    d=Y[i+1]
    E.extend([(b[0]-a[0],X[i][1],X[i+1][1]),(d[0]-c[0],Y[i][1],Y[i+1][1])])
E.sort()

U=[UnionFindNode(i) for i in range(N)]
for cost,v1,v2 in E:
    if not U[v1].is_group(U[v2]):
        U[v1].unite(U[v2])
        ans+=cost
        count+=1
        if count==N:
            break
print(ans)


# Optimized solution
class UnionFindNode(object):
    """
    Union-Find構造
    ノードのグループ併合や、所属グループ判定を高速に処理する
    """

    def __init__(self, group_id, parent=None, value=None):
        self.group_id_ = group_id
        self.parent_ = parent
        self.value = value
        self.size_ = 1

    def __str__(self):
        template = "UnionFindNode(group_id: {}, \n\tparent: {}, value: {}, size: {})"
        return template.format(self.group_id_, self.parent_, self.value, self.size_)

    def is_root(self):
        return not self.parent_

    def root(self):
        parent = self
        while not parent.is_root():
            parent = parent.parent_
            self.parent_ = parent
        return parent

    def find(self):
        parent = self.root()
        return parent.group_id_

    def size(self):
        parent = self.root()
        return parent.size_

    def unite(self, unite_node):
        parent = self.root()
        unite_parent = unite_node.root()

        if parent.group_id_ != unite_parent.group_id_:
            if parent.size() > unite_parent.size():
                unite_parent.parent_ = parent
                parent.size_ = parent.size_ + unite_parent.size_
            else:
                parent.parent_ = unite_parent
                unite_parent.size_ = parent.size_ + unite_parent.size_
    def is_group(self,other):#selfとotherが同じグループに属するか
        return self.root()==other.root()


N=int(eval(input()))
X=[]
Y=[]
for i in range(N):
    x,y=list(map(int,input().split()))
    X.append((x,i))
    Y.append((y,i))
X.sort()
Y.sort()
E=[]
ans=0
count=1
for i in range(N-1):
    a=X[i]
    b=X[i+1]
    c=Y[i]
    d=Y[i+1]
    E.extend([(b[0]-a[0],X[i][1],X[i+1][1]),(d[0]-c[0],Y[i][1],Y[i+1][1])])
E.sort()

U=[UnionFindNode(i) for i in range(N)]
i=0
while(count<N):
    cost,v1,v2=E[i]
    if not U[v1].is_group(U[v2]):
        U[v1].unite(U[v2])
        ans+=cost
        count+=1
    i+=1
print(ans)