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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cassert>
#define rep(i, n) for(i = 0; i < n; i++)
using namespace std;

int n;
char c[4000]; int a[4000];
int wPos[2000];
int bPos[2000];

signed main() {
	int i, j;
	
	cin >> n;
	rep(i, 2 * n) { cin >> c[i] >> a[i]; a[i]--; if (c[i] == 'W') wPos[a[i]] = i; else bPos[a[i]] = i; }
	
	int ans = 0;
	int wCnt = 0, bCnt = 0;
	rep(i, 2 * n) {
		int from; bool isW = false;
		if (wCnt < n && (bCnt == n || wPos[wCnt] < bPos[bCnt])) { from = wPos[wCnt]; isW = true; }
		else { from = bPos[bCnt]; }
		ans += from - i;
		
		//for (j = 0; j < n; j++) cout << "wPos[" << j << "] = " << wPos[j] << endl;
		//for (j = 0; j < n; j++) cout << "bPos[" << j << "] = " << bPos[j] << endl;
		//cout << "from = " << from << " isW = " << isW << endl;
		//cout << endl;
		
		for (j = 0; j < n; j++) if (i <= wPos[j] && wPos[j] < from) wPos[j]++;
		for (j = 0; j < n; j++) if (i <= bPos[j] && bPos[j] < from) bPos[j]++;
		if (isW) { wPos[wCnt++] = i; }
		else { bPos[bCnt++] = i; }
	}
	cout << ans << endl;
	return 0;
}