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 C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<stdint.h>
#include<inttypes.h>

typedef int32_t i32;

#define ALLOC(size,type) ((type*)calloc((size),sizeof(type)))

void add (i32 *bit, i32 x, i32 v) {
  i32 n = bit[0];
  for (i32 i = x; i <= n; i += i & -i) {
    bit[i] += v;
  }
}

i32 get_sum (i32 *bit, i32 x) {
  i32 sum = 0;
  for (i32 i = x; i > 0; i -= i & -i) {
    sum += bit[i];
  }
  return sum;
}

void init_bit (i32 *bit) {
  i32 n = bit[0];
  for (i32 i = 1; i <= n; ++i) {
    bit[i] = 0;
  }
}

void run (void) {
  i32 n;
  scanf ("%" SCNi32, &n);
  i32 *w = ALLOC (n + 1, i32);
  i32 *b = ALLOC (n + 1, i32);
  for (i32 i = 1; i <= 2 * n; ++i) {
    char s[5];
    i32 a;
    scanf ("%s%" SCNi32, s, &a);
    if (s[0] == 'W') {
      w[a] = i;
    } else {
      b[a] = i;
    }
  }
  i32 *bit = ALLOC (2 * n + 1, i32);
  bit[0] = 2 * n;
  i32 *dp = ALLOC (n + 1, i32);
  for (i32 i = 1; i <= n; ++i) {
    add (bit, b[i], 1);
    dp[i] = dp[i - 1] + (i - get_sum (bit, b[i]));
  }
  for (i32 i = 1; i <= n; ++i) {
    init_bit (bit);
    dp[0] = 0;
    for (i32 j = 1; j <= i; ++j) {
      add (bit, w[j], 1);
      dp[0] += j - get_sum (bit, w[j]);
    }
    for (i32 j = 1; j <= n; ++j) {
      add (bit, b[j], 1);
      i32 x = dp[j - 1] + (i + j - get_sum (bit, b[j]));
      i32 y = dp[j] + (i + j - get_sum (bit, w[i]));
      dp[j] = x < y ? x : y;
    }
  }
  printf ("%" PRIi32 "\n", dp[n]);
}

int main (void) {
  run();
  return 0;
}