Problem:
Score : 1200 points

Problem Statement

Snuke is playing with red and blue balls, placing them on a two-dimensional plane.

First, he performed N operations to place red balls. In the i-th of these operations, he placed RC_i red balls at coordinates (RX_i,RY_i).
Then, he performed another N operations to place blue balls. In the i-th of these operations, he placed BC_i blue balls at coordinates (BX_i,BY_i).
The total number of red balls placed and the total number of blue balls placed are equal, that is, \sum_{i=1}^{N} RC_i = \sum_{i=1}^{N} BC_i. Let this value be S.

Snuke will now form S pairs of red and blue balls so that every ball belongs to exactly one pair.
Let us define the score of a pair of a red ball at coordinates (rx, ry) and a blue ball at coordinates (bx, by) as |rx-bx| + |ry-by|.

Snuke wants to maximize the sum of the scores of the pairs.
Help him by finding the maximum possible sum of the scores of the pairs.

Constraints

1 \leq N \leq 1000

0 \leq RX_i,RY_i,BX_i,BY_i \leq 10^9

1 \leq RC_i,BC_i \leq 10

\sum_{i=1}^{N} RC_i = \sum_{i=1}^{N} BC_i

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
RX_1 RY_1 RC_1
RX_2 RY_2 RC_2
\vdots
RX_N RY_N RC_N
BX_1 BY_1 BC_1
BX_2 BY_2 BC_2
\vdots
BX_N BY_N BC_N

Output

Print the maximum possible sum of the scores of the pairs.

Sample Input 1

2
0 0 1
3 2 1
2 2 1
5 0 1

Sample Output 1

8

If we pair the red ball at coordinates (0,0) and the blue ball at coordinates (2,2), the score of this pair is |0-2| + |0-2|=4.
Then, if we pair the red ball at coordinates (3,2) and the blue ball at coordinates (5,0), the score of this pair is |3-5| + |2-0|=4.
Making these two pairs results in the total score of 8, which is the maximum result.

Sample Input 2

3
0 0 1
2 2 1
0 0 2
1 1 1
1 1 1
3 3 2

Sample Output 2

16

Snuke may have performed multiple operations at the same coordinates.

Sample Input 3

10
582463373 690528069 8
621230322 318051944 4
356524296 974059503 6
372751381 111542460 9
392867214 581476334 6
606955458 513028121 5
882201596 791660614 9
250465517 91918758 3
618624774 406956634 6
426294747 736401096 5
974896051 888765942 5
726682138 336960821 3
715144179 82444709 6
599055841 501257806 6
390484433 962747856 4
912334580 219343832 8
570458984 648862300 6
638017635 572157978 10
435958984 585073520 7
445612658 234265014 6

Sample Output 3

45152033546

Solution in Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.lang.RuntimeException
import java.util.*

data class Edge(val from: Int, val to: Int, val capacity: Int, val cost: Long) {
  var resCapacity = capacity
  var resCost = cost
  var revEdge: Edge? = null
}

const val M = 10000000000L

fun main(args: Array<String>) {
  val n = readLine()!!.toInt()
  val edges = mutableListOf<Edge>()
  val s = 2 * n
  val t = 2 * n + 1
  val m1 = 2 * n + 2
  val m2 = 2 * n + 3
  val m3 = 2 * n + 4
  val m4 = 2 * n + 5
  var volume = 0
  repeat(n) {
    val v = it
    val (xInt, yInt, c) = readLine()!!.split(" ").map(String::toInt)
    volume += c
    val x = xInt.toLong()
    val y = yInt.toLong()
    edges.add(Edge(s, v, c, 0))
    edges.add(Edge(v, m1, c, M - (x + y)))
    edges.add(Edge(v, m2, c, M - (x - y)))
    edges.add(Edge(v, m3, c, M - (-x + y)))
    edges.add(Edge(v, m4, c, M - (-x - y)))
  }
  repeat(n) {
    val v = n + it
    val (xInt, yInt, c) = readLine()!!.split(" ").map(String::toInt)
    val x = xInt.toLong()
    val y = yInt.toLong()
    edges.add(Edge(m1, v, c, M - (-x - y)))
    edges.add(Edge(m2, v, c, M - (-x + y)))
    edges.add(Edge(m3, v, c, M - (x - y)))
    edges.add(Edge(m4, v, c, M - (x + y)))
    edges.add(Edge(v, t, c, 0))
  }
  val graph = buildMinCostFlowGraph(2 * n + 6, edges)
  val (flow, cost) = minCostFlow(2 * n + 6, graph, s, t, volume)
  if (flow != volume) throw RuntimeException()
  println(2 * M * volume - cost)
}

fun buildMinCostFlowGraph(n : Int, edges: List<Edge>): List<List<Edge>> {
  val mutableGraph = Array(n, { mutableListOf<Edge>() })
  for (edge in edges) {
    val revEdge = Edge(edge.to, edge.from, 0, -edge.cost)
    edge.revEdge = revEdge
    revEdge.revEdge = edge
    mutableGraph[edge.from].add(edge)
    mutableGraph[revEdge.from].add(revEdge)
  }
  return mutableGraph.map { it.toList() }
}

// Solves a minimum cost flow problem by primal-dual algorithm.
// Complexity: O(B E log V) (B is the volume of flow)
//
// Constraints:
// - There should be no negative cost edges in the initial network.
// - The reverse edges are set in the graph.
//
// Returns a pair of (actual flow volume, cost).
//
// Verified: AOJ-GRL6B, AGC034D
fun minCostFlow(n: Int, graph: List<List<Edge>>,
                s: Int, t: Int, b: Int): Pair<Int, Long> {
  data class DState(val v: Int, val prevEdge: Edge?, val weight: Long)

  val INF = Long.MAX_VALUE / 4
  var flow = 0
  var flowCost = 0L
  while (flow < b) {
    val dists = LongArray(n, { INF })
    val usedEdges = Array<Edge?>(n, { null })

    val pq = PriorityQueue(
        n,
        Comparator<DState> { s1, s2 -> java.lang.Long.compare(s1.weight, s2.weight) })
    pq.add(DState(s, null, 0))
    while (pq.isNotEmpty()) {
      val state = pq.poll()
      if (dists[state.v] < INF) continue
      dists[state.v] = state.weight
      usedEdges[state.v] = state.prevEdge
      if (state.v == t) break
      for (e in graph[state.v]) {
        if (e.resCapacity == 0) continue
        if (dists[e.to] < INF) continue
        val newWeight = state.weight + e.resCost
        if (dists[e.to] <= newWeight) continue
        pq.add(DState(e.to, e, newWeight))
      }
    }

    // There is no path to the sink.
    if (dists[t] == INF) break

    val pathEdges = mutableListOf<Edge>()
    var currentV = t
    while (usedEdges[currentV] != null) {
      pathEdges.add(usedEdges[currentV]!!)
      currentV = usedEdges[currentV]!!.from
    }

    var inc = b - flow
    for (e in pathEdges) {
      inc = Math.min(inc, e.resCapacity)
    }

    flow += inc

    for (e in pathEdges) {
      e.resCapacity -= inc
      e.revEdge!!.resCapacity += inc
      flowCost += inc * e.cost
    }

    for (v in 0 until n) {
      for (e in graph[v]) {
        e.resCost += dists[e.from] - dists[e.to]
      }
    }
  }
  return Pair(flow, flowCost)
}