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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;

int N;
vector<pair<char,int> > vec;

// dp[i][j] := 左から (B, 0), ..., (B, i-1)、(W, 0), ..., (W, j-1) を敷き詰めるまでの最短 swap 数
int dp[2100][2100];

int main() {
    cin >> N;
    vec.resize(N*2);
    for (int i = 0; i < N*2; ++i) {
        cin >> vec[i].first >> vec[i].second;
        --vec[i].second;
    }
    
    // (B, i) の前に (B, iより大きい) + (W, j以上) が何個あるか
    vector<vector<int> > black(N, vector<int>(N+1, 0));
    
    // (W, i) の前に (W, iより大きい) + (B, j以上) が何個あるか
    vector<vector<int> > white(N, vector<int>(N+1, 0));
    
    vector<int> bnum(N, 0);
    vector<int> wnum(N, 0);
    for (int k = 0; k < N*2; ++k) {
        char c = vec[k].first;
        int num = vec[k].second;
        vector<int> wsum(N+1, 0), bsum(N+1, 0);
        for (int l = N; l > 0; --l) {
            bsum[l-1] = bsum[l] + bnum[l-1];
            wsum[l-1] = wsum[l] + wnum[l-1];
        }
        for (int l = N+1; l > 0; --l) {
		  if (c == 'B') black[num][l-1] = wsum[l-1] + bsum[num+1];
		  if (c == 'W') white[num][l-1] = wsum[num+1] + bsum[l-1];
        }
        if (c == 'B')  bnum[num]++;
        else wnum[num]++;
    }
    
    /* DP */
    for (int i = 0; i < 2100; ++i) for (int j = 0; j < 2100; ++j) dp[i][j] = 1<<29;
    dp[0][0] = 0;
    for (int i = 0; i <= N; ++i) {
        for (int j = 0; j <= N; ++j) {
            if (i < N) dp[i+1][j] = min(dp[i+1][j], dp[i][j] + black[i][j]);
            if (j < N) dp[i][j+1] = min(dp[i][j+1], dp[i][j] + white[j][i]);
        }
    }
    cout << dp[N][N] << endl;
}