Problem:
Score : 600 points

Problem Statement

There are 2N balls, N white and N black, arranged in a row. The integers from 1 through N are written on the white balls, one on each ball, and they are also written on the black balls, one on each ball.
The integer written on the i-th ball from the left (1 ≤ i ≤ 2N) is a_i, and the color of this ball is represented by a letter c_i.
c_i = W represents the ball is white; c_i = B represents the ball is black.

Takahashi the human wants to achieve the following objective:

For every pair of integers (i,j) such that 1 ≤ i < j ≤ N, the white ball with i written on it is to the left of the white ball with j written on it.

For every pair of integers (i,j) such that 1 ≤ i < j ≤ N, the black ball with i written on it is to the left of the black ball with j written on it.

In order to achieve this, he can perform the following operation:

Swap two adjacent balls.

Find the minimum number of operations required to achieve the objective.

Constraints

1 ≤ N ≤ 2000

1 ≤ a_i ≤ N

c_i = W or c_i = B.

If i ≠ j, (a_i,c_i) ≠ (a_j,c_j).

Input

Input is given from Standard Input in the following format:

N
c_1 a_1
c_2 a_2
:
c_{2N} a_{2N}

Output

Print the minimum number of operations required to achieve the objective.

Sample Input 1

3
B 1
W 2
B 3
W 1
W 3
B 2

Sample Output 1

4

The objective can be achieved in four operations, for example, as follows:

Swap the black 3 and white 1.

Swap the white 1 and white 2.

Swap the black 3 and white 3.

Swap the black 3 and black 2.

Sample Input 2

4
B 4
W 4
B 3
W 3
B 2
W 2
B 1
W 1

Sample Output 2

18

Sample Input 3

9
W 3
B 1
B 4
W 1
B 5
W 9
W 2
B 6
W 5
B 3
W 8
B 9
W 7
B 2
B 8
W 4
W 6
B 7

Sample Output 3

41

Solution in Kotlin (1.0.0) (judge verdict: Accepted)

data class Ball(val c: Char, val v: Int)

const val INF = Int.MAX_VALUE / 4

fun main(args: Array<String>) {
  val n = readLine()!!.toInt()
  val balls = Array( 2 * n, {
    val (cString, vString) = readLine()!!.split(" ")
    val c = cString[0]
    val v = vString.toInt()
    Ball(c, v)
  })
      .toMutableList()

  val whiteValueToIndex = IntArray(n + 1, { -1 })
  val blackValueToIndex = IntArray(n + 1, { -1 })
  for (i in 0 until 2 * n) {
    if (balls[i].c == 'W') {
      whiteValueToIndex[balls[i].v] = i
    } else {
      blackValueToIndex[balls[i].v] = i
    }
  }

  val whiteLeft = Array(2 * n, { BooleanArray(n + 1) })
  val blackLeft = Array(2 * n, { BooleanArray(n + 1) })
  val countWhiteLE = Array(2 * n, { IntArray(n + 1) })
  val countBlackLE = Array(2 * n, { IntArray(n + 1) })
  for (i in 0 until 2 * n) {
    if (balls[i].c == 'W') {
      for (j in i + 1 until 2 * n) {
        whiteLeft[j][balls[i].v] = true
      }
    } else {
      for (j in i + 1 until 2 * n) {
        blackLeft[j][balls[i].v] = true
      }
    }
  }
  for (i in 0 until 2 * n) {
    for (k in 1..n) {
      countWhiteLE[i][k] = countWhiteLE[i][k - 1] + if (whiteLeft[i][k]) 1 else 0
      countBlackLE[i][k] = countBlackLE[i][k - 1] + if (blackLeft[i][k]) 1 else 0
    }
  }

  val dp = Array(n + 1, { IntArray(n + 1, { INF }) })
  dp[0][0] = 0

  for (w in 0..n) {
    for (b in 0..n) {
      // w White and b Black are used now.

      // w + 1 White
      if (w < n) {
        val index = whiteValueToIndex[w + 1]
        // val index = balls.indexOfFirst { it.c == 'W' && it.v == w + 1 }
        val cost = index - countWhiteLE[index][w] - countBlackLE[index][b]
        dp[w + 1][b] = Math.min(dp[w + 1][b], dp[w][b] + cost)
      }

      // b + 1 Black
      if (b < n) {
        val index = blackValueToIndex[b + 1]
        // val index = balls.indexOfFirst { it.c == 'B' && it.v == b + 1 }
        val cost = index - countWhiteLE[index][w] - countBlackLE[index][b]
        dp[w][b + 1] = Math.min(dp[w][b + 1], dp[w][b] + cost)
      }
    }
  }
  println(dp[n][n])
}