output_description: Print the minimum necessary amount of money in order to build roads so that it
  will be possible to travel between every pair of towns by traversing roads.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      x_1 y_1
      x_2 y_2
      :
      x_N y_N
attempt: import heapq
  
  
  class Graph:
      def __init__(self, v, edgelist, w_v=None, directed=False):
          super().__init__()
          self.v = v
          self.w_e = [{} for _ in [0] * self.v]
          self.neighbor = [[] for _ in [0] * self.v]
          self.w_v = w_v
          self.directed = directed
  
          for i, j, w in edgelist:
              self.w_e[i][j] = w
              self.neighbor[i].append(j)
  
      def dijkstra(self, v_n):
          d = [float("inf")] * self.v
          d[v_n] = 0
          prev = [-1] * self.v
  
          queue = []
          for i, d_i in enumerate(d):
              heapq.heappush(queue, (d_i, i))
  
          while len(queue) > 0:
              d_u, u = queue.pop()
              if d[u] < d_u:
                  continue
              for v in self.neighbor[u]:
                  alt = d[u] + self.w_e[u][v]
                  if d[v] > alt:
                      d[v] = alt
                      prev[v] = u
                      heapq.heappush(queue, (alt, v))
  
          return d, prev
  
      def warshallFloyd(self):
          d = [[10**18] * self.v for _ in [0] * self.v]
          for i in range(self.v):
              d[i][i] = 0
  
          for i in range(self.v):
              for j in self.neighbor[i]:
                  d[i][j] = self.w_e[i][j]
  
          for i in range(self.v):
              for j in self.neighbor[i]:
                  d[i][j] = self.w_e[i][j]
  
          for k in range(self.v):
              for i in range(self.v):
                  for j in range(self.v):
                      check = d[i][k] + d[k][j]
                      if d[i][j] > check:
                          d[i][j] = check
          return d
  
      def prim(self):
          gb = GraphBuilder(self.v, self.directed)
  
          queue = []
          for i, w in self.w_e[0].items():
              heapq.heappush(queue, (w, 0, i))
          rest = [True] * self.v
          rest[0] = False
  
          while len(queue) > 0:
              w, i, j = heapq.heappop(queue)
              if rest[j]:
                  gb.addEdge(i, j, w)
                  rest[j] = False
                  for k, w in self.w_e[j].items():
                      heapq.heappush(queue, (w, j, k))
  
          return gb
  
  
  class Tree:
      def __init__(self, v, e):
          pass
  
  
  class GraphBuilder:
      def __init__(self, v, directed=False):
          self.v = v
          self.directed = directed
          self.edge = []
  
      def addEdge(self, i, j, w=1):
          if not self.directed:
              self.edge.append((j, i, w))
          self.edge.append((i, j, w))
  
      def addEdges(self, edgelist, weight=True):
          if weight:
              if self.directed:
                  for i, j, w in edgelist:
                      self.edge.append((i, j, w))
              else:
                  for i, j, w in edgelist:
                      self.edge.append((i, j, w))
                      self.edge.append((j, i, w))
          else:
              if self.directed:
                  for i, j, w in edgelist:
                      self.edge.append((i, j, 1))
              else:
                  for i, j, w in edgelist:
                      self.edge.append((i, j, 1))
                      self.edge.append((j, i, 1))
  
      def addAdjMat(self, mat):
          for i, mat_i in enumerate(mat):
              for j, w in enumerate(mat_i):
                  self.edge.append((i, j, w))
  
      def buildTree(self):
          pass
  
      def buildGraph(self):
          return Graph(self.v, self.edge, directed=self.directed)
  
  
  def main():
  
      n = int(input())
      edge = []
      for i in range(n):
          x, y = list(map(int, input().split()))
          edge.append((x, y, i))
  
      xs = sorted(edge, key=lambda a: a[0])
      ys = sorted(edge, key=lambda a: a[1])
  
      edge = []
  
      for i in range(n - 1):
          x1, y1, p1 = xs[i]
          x2, y2, p2 = xs[i + 1]
          edge.append((p1, p2, min(x2 - x1, y2 - y1)))
  
          x1, y1, p1 = ys[i]
          x2, y2, p2 = ys[i + 1]
          edge.append((p1, p2, y2 - y1))
  
      gb = GraphBuilder(n)
      gb.addEdges(edge)
  
      mint = gb.buildGraph().prim()
      print(mint.edge)
      ans = 0
      for a in mint.edge:
          ans += a[2]
      print(ans // 2)
  
  
  if __name__ == "__main__":
      main()
  
problem_description: Statement
  
  There are N towns on a plane. The i-th town is located at the coordinates
  (x_i,y_i). There may be more than one town at the same coordinates.
  
  You can build a road between two towns at coordinates (a,b) and (c,d) for a
  cost of min(|a-c|,|b-d|) yen (the currency of Japan). It is not possible to
  build other types of roads.
  
  Your objective is to build roads so that it will be possible to travel between
  every pair of towns by traversing roads. At least how much money is necessary
  to achieve this?
samples: [{"input": "3\n    1 5\n    3 9\n    7 8", "output": "3\n    \n\nBuild a road between Towns 1 and 2, and another between Towns 2 and 3. The\ntotal cost is 2+1=3 yen.\n\n* * *"}, {"input": "6\n    8 3\n    4 9\n    12 19\n    18 1\n    13 5\n    7 6", "output": "8"}]