// Original C++ solution (slower):
#include <bits/stdc++.h>

#define rep(i,n) for (int i = (0); i < (n); ++i)

#define rrep(i,n) for(int i = 1; i <= (n); ++i)

#define drep(i,n) for(int i = (n)-1; i >= 0; --i)

#define srep(i,s,t) for (int i = s; i < t; ++i)

#define rng(x) (x).begin(),(x).end()

#define rrng(x) (x).rbegin(),(x).rend()

#define limit(x,l,r) max(l,min(x,r))

#define lims(x,l,r) (x = max(l,min(x,r)))

#define isin(x,l,r) ((l) <= (x) && (x) < (r))

#define show(x) cout << #x << " = " << (x) << endl

#define show2(x,y) cout << #x << " = " << (x) << ", " << #y << " = " << (y) << endl

#define show3(x,y,z) cout << #x << " = " << (x) << ", " << #y << " = " << (y) <<  ", " << #z << " = " << (z) << endl

#define showv(v) rep(i,v.size()) printf("%d%c", v[i], i==v.size()-1?'\n':' ')

#define showv2(v) rep(j,v.size()) showv(v[j])

#define showt(t,n) rep(i,n) printf("%d%c", t[i], i==n-1?'\n':' ')

#define showt2(t,r,c) rep(j,r) showt(t[j],c)

#define showvp(p) rep(i,p.size()) printf("%d %d\n", p[i].first, p[i].second);

#define printv(v) rep(i,v.size()) printf("%d\n", v[i])

#define printt(t,n) rep(i,n) printf("%d\n", t[i])

#define incl(v,x) find(rng(v),x)!=v.end()

#define incls(s,c) s.find(c)!=string::npos

#define fi first

#define se second

#define pb push_back

#define eb emplace_back

#define sz(x) (int)(x).size()

#define pcnt __builtin_popcountll

#define bit(n,k) ((n>>k)&1) // nのk bit目

#define bn(x) ((1<<x)-1)

#define dup(x,y) (((x)+(y)-1)/(y))

#define newline puts("")

#define uni(x) x.erase(unique(rng(x)),x.end())

#define SP << " " <<

#define v(T) vector<T>

#define vv(T) v(v(T))

using namespace std;

using ll = long long;

using ld = long double;

using vi = vector<int>;

using vvi = vector<vi>;

using vl = vector<ll>;

using vvl = vector<vl>;

using vs = vector<string>;

using P = pair<ll, ll>;

using T = tuple<ll, ll, ll>;

using vp = vector<P>;

using vt = vector<T>;

const int mod = 1000000007;

const double EPS = 1e-9;

//const long double EPS = 1e-14;

const int INF = (1<<30)-1;

const ll LINF = (1LL<<62)-1;

#define dame { puts("No"); return 0;}

#define yn {puts("Yes");}else{puts("No");}

inline int in() { int x; scanf("%d",&x); return x;}

inline ll lin() { ll x; scanf("%lld",&x); return x;}

inline char chin() { char x; scanf("%c",&x); return x;}

inline string stin() { string x; cin >> x; return x;}

inline double din() { double x; scanf("%lf",&x); return x;}

//template<class T = int> inline T in() { T x; cin >> x; return (x);}

template<typename T>inline ll suma(const v(T)& a) { ll res(0); for (auto&& x : a) res += x; return res;}

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

char itoa(int n) { return n + '0';}

ll gcd(ll a, ll b) { return b ? gcd(b,a%b) : a;}

ll lcm(ll a, ll b) { return a/gcd(a,b)*b;}



const ll MAX_V = 200005;

const ll MAX_E = 200005;

ll V; // 頂点数

ll E; // 辺の本数



struct edge{ ll u, v, cost;};

edge es[MAX_E];



struct UnionFind {

  vector<ll> d;

  UnionFind(ll n): d(n,-1) {}

  ll find(ll x) {

    if (d[x] < 0) return x;

    return d[x] = find(d[x]);

  }

  bool unite(ll x, ll y) {

    x = find(x); y = find(y);

    if (x == y) return false;

    if (d[x] > d[y]) swap(x,y); // -d[x] < -d[y]

    d[x] += d[y];

    d[y] = x;

    return true;

  }

  bool same(ll x, ll y) { return find(x) == find(y);}

  ll size(ll x) { return -d[find(x)];}

};



// クラスカル法: 最小全域木のコストの合計値を返す O(|E|log|V|)

ll kruskal() {

  auto comp = [](const edge& e1, const edge& e2) {

    return e1.cost < e2.cost;

  };

  sort(es, es + E, comp);

  UnionFind uf(MAX_V);

  ll res = 0;

  for (ll i = 0; i < E; ++i) {

    edge e = es[i];

    if (!uf.same(e.u, e.v)) {

      uf.unite(e.u, e.v);

      res += e.cost;

    }

  }

  return res;

}



int main () {

  ll n;

  cin >> n;

  vt p(n);

  rep(i,n) {

    ll x, y;

    cin >> x >> y;

    x--, y--;

    p[i] = make_tuple(x,y,i);

  }

  auto q = p;

  sort(rng(p));



  auto comp_y = [&](T a, T b) {

    if (get<1>(a) != get<1>(b)) return get<1>(a) < get<1>(b);

    else return get<0>(a) < get<0>(b);

  };

  sort(rng(q),comp_y);



  V = 2*n;

  E = 2*(n-1);

  rep(i,n-1) {

    es[2*i] = (edge){get<2>(p[i]), get<2>(p[i+1]), abs(get<0>(p[i])-get<0>(p[i+1]))};

    es[2*i+1] = (edge){get<2>(q[i]), get<2>(q[i+1]), abs(get<1>(q[i])-get<1>(q[i+1]))};

  }

  ll ans = kruskal();

  cout << ans << endl;

  return 0;

}

// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>

#define rep(i,n) for (int i = (0); i < (n); ++i)

#define rrep(i,n) for(int i = 1; i <= (n); ++i)

#define drep(i,n) for(int i = (n)-1; i >= 0; --i)

#define srep(i,s,t) for (int i = s; i < t; ++i)

#define rng(x) (x).begin(),(x).end()

#define rrng(x) (x).rbegin(),(x).rend()

#define limit(x,l,r) max(l,min(x,r))

#define lims(x,l,r) (x = max(l,min(x,r)))

#define isin(x,l,r) ((l) <= (x) && (x) < (r))

#define show(x) cout << #x << " = " << (x) << endl

#define show2(x,y) cout << #x << " = " << (x) << ", " << #y << " = " << (y) << endl

#define show3(x,y,z) cout << #x << " = " << (x) << ", " << #y << " = " << (y) <<  ", " << #z << " = " << (z) << endl

#define showv(v) rep(i,v.size()) printf("%d%c", v[i], i==v.size()-1?'\n':' ')

#define showv2(v) rep(j,v.size()) showv(v[j])

#define showt(t,n) rep(i,n) printf("%d%c", t[i], i==n-1?'\n':' ')

#define showt2(t,r,c) rep(j,r) showt(t[j],c)

#define showvp(p) rep(i,p.size()) printf("%d %d\n", p[i].first, p[i].second);

#define printv(v) rep(i,v.size()) printf("%d\n", v[i])

#define printt(t,n) rep(i,n) printf("%d\n", t[i])

#define incl(v,x) find(rng(v),x)!=v.end()

#define incls(s,c) s.find(c)!=string::npos

#define fi first

#define se second

#define pb push_back

#define eb emplace_back

#define sz(x) (int)(x).size()

#define pcnt __builtin_popcountll

#define bit(n,k) ((n>>k)&1) // nのk bit目

#define bn(x) ((1<<x)-1)

#define dup(x,y) (((x)+(y)-1)/(y))

#define newline puts("")

#define uni(x) x.erase(unique(rng(x)),x.end())

#define SP << " " <<

#define v(T) vector<T>

#define vv(T) v(v(T))

using namespace std;

using ll = long long;

using ld = long double;

using vi = vector<int>;

using vvi = vector<vi>;

using vl = vector<ll>;

using vvl = vector<vl>;

using vs = vector<string>;

using P = pair<int, int>;

using T = tuple<int, int, int>;

using vp = vector<P>;

using vt = vector<T>;

const int mod = 1000000007;

const double EPS = 1e-9;

//const long double EPS = 1e-14;

const int INF = (1<<30)-1;

const ll LINF = (1LL<<62)-1;

#define dame { puts("No"); return 0;}

#define yn {puts("Yes");}else{puts("No");}

inline int in() { int x; scanf("%d",&x); return x;}

inline ll lin() { ll x; scanf("%lld",&x); return x;}

inline char chin() { char x; scanf("%c",&x); return x;}

inline string stin() { string x; cin >> x; return x;}

inline double din() { double x; scanf("%lf",&x); return x;}

//template<class T = int> inline T in() { T x; cin >> x; return (x);}

template<typename T>inline ll suma(const v(T)& a) { ll res(0); for (auto&& x : a) res += x; return res;}

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

char itoa(int n) { return n + '0';}

ll gcd(ll a, ll b) { return b ? gcd(b,a%b) : a;}

ll lcm(ll a, ll b) { return a/gcd(a,b)*b;}



const int MAX_V = 200005;

const int MAX_E = 200005;

int V; // 頂点数

int E; // 辺の本数



struct edge{ int u, v, cost;};

edge es[MAX_E];



struct UnionFind {

  vector<int> d;

  UnionFind(int n): d(n,-1) {}

  int find(int x) {

    if (d[x] < 0) return x;

    return d[x] = find(d[x]);

  }

  bool unite(int x, int y) {

    x = find(x); y = find(y);

    if (x == y) return false;

    if (d[x] > d[y]) swap(x,y); // -d[x] < -d[y]

    d[x] += d[y];

    d[y] = x;

    return true;

  }

  bool same(int x, int y) { return find(x) == find(y);}

  int size(int x) { return -d[find(x)];}

};



// クラスカル法: 最小全域木のコストの合計値を返す O(|E|log|V|)

int kruskal() {

  auto comp = [](const edge& e1, const edge& e2) {

    return e1.cost < e2.cost;

  };

  sort(es, es + E, comp);

  UnionFind uf(MAX_V);

  int res = 0;

  for (int i = 0; i < E; ++i) {

    edge e = es[i];

    if (!uf.same(e.u, e.v)) {

      uf.unite(e.u, e.v);

      res += e.cost;

    }

  }

  return res;

}



int main () {

  int n;

  cin >> n;

  vt p(n);

  rep(i,n) {

    int x, y;

    cin >> x >> y;

    x--, y--;

    p[i] = make_tuple(x,y,i);

  }

  auto q = p;

  sort(rng(p));



  auto comp_y = [&](T a, T b) {

    if (get<1>(a) != get<1>(b)) return get<1>(a) < get<1>(b);

    else return get<0>(a) < get<0>(b);

  };

  sort(rng(q),comp_y);



  V = 2*n;

  E = 2*(n-1);

  rep(i,n-1) {

    es[2*i] = (edge){get<2>(p[i]), get<2>(p[i+1]), abs(get<0>(p[i])-get<0>(p[i+1]))};

    es[2*i+1] = (edge){get<2>(q[i]), get<2>(q[i+1]), abs(get<1>(q[i])-get<1>(q[i+1]))};

  }

  int ans = kruskal();

  cout << ans << endl;

  return 0;

}

// Fastest known solution for this problem: