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.

Examples

Input

3
B 1
W 2
B 3
W 1
W 3
B 2


Output

4


Input

4
B 4
W 4
B 3
W 3
B 2
W 2
B 1
W 1


Output

18


Input

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


Output

41

#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <map>
#include <unordered_map>
#include <set>
#include <functional>
using namespace std;

typedef long long ll;

pair<char, int> c[4004];
int pb[2004], pw[2004];
int b[4004][2004], w[4004][2004];
int dp[2004][2004];

int main() {
	int n; cin >> n;
	for (int i = 1; i <= 2 * n; i++) {
		cin >> c[i].first >> c[i].second;
		(c[i].first == 'B' ? pb : pw)[c[i].second] = i;
	}

	for (int i = 2 * n; i >= 1; i--) {
		for (int j = 1; j <= n + 1; j++) {
			b[i][j] = b[i + 1][j] + (c[i].first == 'B' && c[i].second < j);
			w[i][j] = w[i + 1][j] + (c[i].first == 'W' && c[i].second < j);
		}
	}

	for (int i = 0; i <= n; i++) {
		for (int j = 0; j <= n; j++) {
			if (i > 0 && j > 0) dp[i][j] = min(dp[i - 1][j] + b[pb[i]][i] + w[pb[i]][j + 1], dp[i][j - 1] + b[pw[j]][i + 1] + w[pw[j]][j]);
			else if (i > 0) dp[i][j] = dp[i - 1][j] + b[pb[i]][i] + w[pb[i]][j + 1];
			else if (j > 0) dp[i][j] = dp[i][j - 1] + b[pw[j]][i + 1] + w[pw[j]][j];
		}
	}

	cout << dp[n][n] << endl;
	return 0;
}