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 Rust (1.15.1) (judge verdict: Accepted)

#![allow(non_snake_case)]
#![allow(unused_variables)]
#![allow(dead_code)]

fn main() {
    let N: usize = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.trim().parse().unwrap()
    };
    let (c, a): (Vec<char>, Vec<usize>) = {
        let (mut c, mut a) = (vec![], vec![]);
        for _ in 0..(2 * N) {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            c.push(iter.next().unwrap().chars().collect::<Vec<_>>()[0]);
            a.push(iter.next().unwrap().parse().unwrap());
        }
        (c, a)
    };
    let mut W = vec![0; N + 1];
    let mut B = vec![0; N + 1];
    for i in 0..(2 * N) {
        if c[i] == 'W' {
            W[a[i]] = i;
        }
        if c[i] == 'B' {
            B[a[i]] = i;
        }
    }

    const INF: usize = 1_000_000_000_000_000;
    let mut dp = vec![vec![INF; N + 1]; N + 1];
    dp[0][0] = 0;
    let mut x = vec![vec![0; N + 1]; N + 1];
    for i in 0..(N + 1) {
        x[i][0] = (0..W[i]).filter(|&k| c[k] == 'W' && a[k] > i).count()
            + (0..W[i]).filter(|&k| c[k] == 'B').count();
        for j in 1..(N + 1) {
            x[i][j] = if B[j] < W[i] {
                x[i][j - 1] - 1
            } else {
                x[i][j - 1]
            };
        }
    }
    let mut y = vec![vec![0; N + 1]; N + 1];
    for j in 0..(N + 1) {
        y[0][j] = (0..B[j]).filter(|&k| c[k] == 'B' && a[k] > j).count()
            + (0..B[j]).filter(|&k| c[k] == 'W').count();
        for i in 1..(N + 1) {
            y[i][j] = if W[i] < B[j] {
                y[i - 1][j] - 1
            } else {
                y[i - 1][j]
            };
        }
    }
    for i in 0..(N + 1) {
        for j in 0..(N + 1) {
            dp[i][j] = if i == 0 && j == 0 {
                0
            } else if i == 0 {
                dp[i][j - 1] + y[i][j]
            } else if j == 0 {
                dp[i - 1][j] + x[i][j]
            } else {
                std::cmp::min(dp[i][j - 1] + y[i][j], dp[i - 1][j] + x[i][j])
            };
        }
    }
    let ans = dp[N][N];

    println!("{}", ans);
}