#include <bits/stdc++.h>
using namespace std;
const long double INF = 1e18;
const long double EPS = 1e-12;
bool equals(long double a, long double b) { return fabs(a - b) < EPS; }
struct edge {
  int u, v;
  long long cap;
  long double cost;
  edge(int _u, int _v, long long _cap, long double _cost) {
    u = _u;
    v = _v;
    cap = _cap;
    cost = _cost;
  }
};
struct min_cost_flow {
 public:
  int n, s, t;
  long long flow;
  long double cost;
  vector<vector<int> > graph;
  vector<edge> e;
  vector<long double> dist;
  vector<int> parent;
  min_cost_flow(int _n) {
    n = _n;
    graph.assign(n, vector<int>());
  }
  void add_edge(int u, int v, long long cap, long double _cost,
                bool directed = true) {
    graph[u].push_back(e.size());
    e.push_back(edge(u, v, cap, _cost));
    graph[v].push_back(e.size());
    e.push_back(edge(v, u, 0, -_cost));
    if (!directed) {
      add_edge(v, u, cap, cost, true);
    }
  }
  pair<long long, long double> get_min_cost_flow(int _s, int _t) {
    s = _s;
    t = _t;
    flow = 0, cost = 0;
    while (SPFA()) {
      flow += send_flow(t, 1LL << 62);
    }
    return make_pair(flow, cost);
  }

 private:
  bool SPFA() {
    parent.assign(n, -1);
    dist.assign(n, INF);
    dist[s] = 0;
    vector<int> queuetime(n, 0);
    queuetime[s] = 1;
    vector<bool> inqueue(n, 0);
    inqueue[s] = true;
    queue<int> q;
    q.push(s);
    bool negativecycle = false;
    while (!q.empty() && !negativecycle) {
      int u = q.front();
      q.pop();
      inqueue[u] = false;
      for (int i = 0; i < (int)graph[u].size(); i++) {
        int eIdx = graph[u][i];
        int v = e[eIdx].v;
        long double w = e[eIdx].cost;
        long long cap = e[eIdx].cap;
        if (dist[u] + w + EPS < dist[v] && cap > 0) {
          dist[v] = dist[u] + w;
          parent[v] = eIdx;
          if (!inqueue[v]) {
            q.push(v);
            queuetime[v]++;
            inqueue[v] = true;
            if (queuetime[v] == n + 2) {
              negativecycle = true;
              break;
            }
          }
        }
      }
    }
    return !equals(dist[t], INF);
  }
  long long send_flow(int v, long long cur_flow) {
    if (parent[v] == -1) {
      return cur_flow;
    }
    int eIdx = parent[v];
    int u = e[eIdx].u;
    long double w = e[eIdx].cost;
    long long f = send_flow(u, min(cur_flow, e[eIdx].cap));
    cost += f * w;
    e[eIdx].cap -= f;
    e[eIdx ^ 1].cap += f;
    return f;
  }
};
long double dis(pair<int, int> a, pair<int, int> b) {
  int dx = a.first - b.first;
  int dy = a.second - b.second;
  return sqrt(dx * dx + dy * dy);
}
const int MAX = 1e6 + 6;
int cnt[MAX];
int n;
int main() {
  scanf("%d", &n);
  vector<pair<int, int> > v;
  int max_y = 0;
  for (int i = int(1); i < int(n + 1); i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    cnt[y + 1000]++;
    max_y = max(max_y, y + 1000);
    v.emplace_back(y, x);
  }
  sort(v.begin(), v.end(), greater<pair<int, int> >());
  if (cnt[max_y] != 1) {
    puts("-1");
    return 0;
  }
  min_cost_flow mcmf(3 * n);
  int source = 2 * n, sink = 2 * n + 1;
  for (int i = int(0); i < int(n); i++) {
    if (i) {
      mcmf.add_edge(source, i, 1, 0);
    }
    mcmf.add_edge(i + n, sink, 2, 0);
  }
  for (int i = int(1); i < int(v.size()); i++) {
    for (int j = int(0); j < int(i); j++) {
      if (v[j].first > v[i].first) {
        mcmf.add_edge(i, j + n, 1, dis(v[i], v[j]));
      }
    }
  }
  pair<long long, long double> p = mcmf.get_min_cost_flow(source, sink);
  if (p.first != n - 1) {
    puts("-1");
  } else {
    cout << fixed << setprecision(15) << p.second << endl;
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

A root tree is a directed acyclic graph that contains one node (root), from which there is exactly one path to any other node.

A root tree is binary if each node has at most two outgoing arcs.

When a binary tree is painted on the plane, all arcs should be directed from top to bottom. That is, each arc going from u to v must meet the condition yu > yv.

You've been given the coordinates of all tree nodes. Your task is to connect these nodes by arcs so as to get the binary root tree and make the total length of the arcs minimum. All arcs of the built tree must be directed from top to bottom.

Input

The first line contains a single integer n (2 ≤ n ≤ 400) — the number of nodes in the tree. Then follow n lines, two integers per line: xi, yi (|xi|, |yi| ≤ 103) — coordinates of the nodes. It is guaranteed that all points are distinct.

Output

If it is impossible to build a binary root tree on the given points, print "-1". Otherwise, print a single real number — the total length of the arcs in the minimum binary tree. The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6. 

Examples

Input

3
0 0
1 0
2 1


Output

3.650281539872885


Input

4
0 0
1 0
2 1
2 0


Output

-1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long double INF = 1e18;
const long double EPS = 1e-12;
bool equals(long double a, long double b) { return fabs(a - b) < EPS; }
struct edge {
  int u, v;
  long long cap;
  long double cost;
  edge(int _u, int _v, long long _cap, long double _cost) {
    u = _u;
    v = _v;
    cap = _cap;
    cost = _cost;
  }
};
struct min_cost_flow {
 public:
  int n, s, t;
  long long flow;
  long double cost;
  vector<vector<int> > graph;
  vector<edge> e;
  vector<long double> dist;
  vector<int> parent;
  min_cost_flow(int _n) {
    n = _n;
    graph.assign(n, vector<int>());
  }
  void add_edge(int u, int v, long long cap, long double _cost,
                bool directed = true) {
    graph[u].push_back(e.size());
    e.push_back(edge(u, v, cap, _cost));
    graph[v].push_back(e.size());
    e.push_back(edge(v, u, 0, -_cost));
    if (!directed) {
      add_edge(v, u, cap, cost, true);
    }
  }
  pair<long long, long double> get_min_cost_flow(int _s, int _t) {
    s = _s;
    t = _t;
    flow = 0, cost = 0;
    while (SPFA()) {
      flow += send_flow(t, 1LL << 62);
    }
    return make_pair(flow, cost);
  }

 private:
  bool SPFA() {
    parent.assign(n, -1);
    dist.assign(n, INF);
    dist[s] = 0;
    vector<int> queuetime(n, 0);
    queuetime[s] = 1;
    vector<bool> inqueue(n, 0);
    inqueue[s] = true;
    queue<int> q;
    q.push(s);
    bool negativecycle = false;
    while (!q.empty() && !negativecycle) {
      int u = q.front();
      q.pop();
      inqueue[u] = false;
      for (int i = 0; i < (int)graph[u].size(); i++) {
        int eIdx = graph[u][i];
        int v = e[eIdx].v;
        long double w = e[eIdx].cost;
        long long cap = e[eIdx].cap;
        if (dist[u] + w + EPS < dist[v] && cap > 0) {
          dist[v] = dist[u] + w;
          parent[v] = eIdx;
          if (!inqueue[v]) {
            q.push(v);
            queuetime[v]++;
            inqueue[v] = true;
            if (queuetime[v] == n + 2) {
              negativecycle = true;
              break;
            }
          }
        }
      }
    }
    return !equals(dist[t], INF);
  }
  long long send_flow(int v, long long cur_flow) {
    if (parent[v] == -1) {
      return cur_flow;
    }
    int eIdx = parent[v];
    int u = e[eIdx].u;
    long double w = e[eIdx].cost;
    long long f = send_flow(u, min(cur_flow, e[eIdx].cap));
    cost += f * w;
    e[eIdx].cap -= f;
    e[eIdx ^ 1].cap += f;
    return f;
  }
};
long double dis(pair<int, int> a, pair<int, int> b) {
  int dx = a.first - b.first;
  int dy = a.second - b.second;
  return sqrt(dx * dx + dy * dy);
}
const int MAX = 1e6 + 6;
int cnt[MAX];
int n;
int main() {
  scanf("%d", &n);
  vector<pair<int, int> > v;
  int max_y = 0;
  for (int i = int(1); i < int(n + 1); i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    cnt[y + 1000]++;
    max_y = max(max_y, y + 1000);
    v.emplace_back(y, x);
  }
  sort(v.begin(), v.end(), greater<pair<int, int> >());
  if (cnt[max_y] != 1) {
    puts("-1");
    return 0;
  }
  min_cost_flow mcmf(3 * n);
  int source = 2 * n, sink = 2 * n + 1;
  for (int i = int(0); i < int(n); i++) {
    if (i) {
      mcmf.add_edge(source, i, 1, 0);
    }
    mcmf.add_edge(i + n, sink, 2, 0);
  }
  for (int i = int(1); i < int(v.size()); i++) {
    for (int j = int(0); j < int(i); j++) {
      if (v[j].first > v[i].first) {
        mcmf.add_edge(i, j + n, 1, dis(v[i], v[j]));
      }
    }
  }
  pair<long long, long double> p = mcmf.get_min_cost_flow(source, sink);
  if (p.first != n - 1) {
    puts("-1");
  } else {
    cout << fixed << setprecision(15) << p.second << endl;
  }
  return 0;
}

```