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: Runtime Error)

#include <bits/stdc++.h>

#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define rep(i,n) FOR(i,0,n)
#define RFOR(i,a,b) for(int i=(a)-1;i>=(b);i--)
#define rrep(i,n) RFOR(i,n,0)

using namespace std;

typedef long long ll;
typedef unsigned long long ull;

const int INF = 1e9;

// Binary Indexed Tree
const int max_n  = 5000;

class BinaryIndexedTree{

private:
	int n;
	ll bit[max_n];

public:
	BinaryIndexedTree(int n){
		this -> n = n;
	}

	ll sum(int i){
	  ll s = 0;
	  while(i > 0){
	    s += bit[i];
	    i -= i & (-i);
	  }
	  return s;
	}

	void add(int i, ll x){
	  while(i <= n){
	    bit[i] += x;
	    i += i & (-i);
	  }
	}

	void show(){
		rep(i,n) cout << bit[i] << ",";
		cout << endl;
	}
};

// あわない…

int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	int n;
	cin >> n;

	int bl[2][n+1];
	rep(i, 2*n){
		string c;
		int x;

		cin >> c >> x;
		bl[(c == "W" ? 1 : 0)][x] = i+1;
	}

	// costb[i][j] := 黒がi個、白がj個置かれてある状態で黒のボールを置くとき、そのボールの転倒数
	// これは先においてある黒、白のボールの並びによらない
	ll costb[n+1][n+1];
	ll costw[n+1][n+1];

	int bcnt = 0;
	int wcnt = 0;

	BinaryIndexedTree bit = BinaryIndexedTree(2*n+1);

	rep(i, n+1){
		if(i > 0) {bit.add(bl[0][i], 1); }
		
		rep(j, n+1){
			if(j > 0) {bit.add(bl[1][j], 1);}
			// cost[i][j] = (すでにおいてあるボール) -　(転倒しないボール=最初からそのボールより左においてあるボール)
			if(i < n+1) costb[i][j] = i + j - bit.sum(bl[0][i+1]);
			if(j < n+1) costw[i][j] = i + j - bit.sum(bl[1][j+1]);
		}

		// bitの状態を戻す
		rep(j, n){ bit.add(bl[1][j+1], -1);}
	}

	// rep(i,n+1) rep(j,n+1) cout << i << "," << j << "," << costb[i][j] << "," << costw[i][j] << endl;

	// dp[i][j] := 黒をi個、白をj個置いたときの各ボールの転倒数の合計の最小値
	ll dp[n+1][n+1];
	rep(i, n+1) rep(j, n+1) dp[i][j] = INF;
	dp[0][0] = 0;
	dp[0][1] = costw[0][0];
	dp[1][0] = costb[0][0];

	rep(i, n){
		rep(j, n){
			dp[i+1][j+1] = min(dp[i][j+1] + costb[i][j+1], dp[i+1][j] + costw[i+1][j]);
		}
	}

	cout << dp[n][n] << endl;	
}