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 <cstdio>
#include <algorithm>

using namespace std;

const int inf = 1e9;
const int maxN = 4005;
int p[2][maxN];
int cost[2][maxN][maxN];
int dp[maxN][maxN];

int main() {
	int N; cin >> N; 
	for(int i = 1; i <= 2 * N; ++i) {
		char c; int x; cin >> c >> x;
		p[c == 'W'][x] = i;
	}
	for(int k = 0; k < 2; ++k) {
		for(int i = 1; i <= N; ++i) {
			for(int j = 1; j < i; ++j) cost[k][i][0] += (p[k][j] > p[k][i]);
			for(int j = 1; j <= N; ++j) cost[k][i][j] = cost[k][i][j - 1] + (p[1 - k][j] > p[k][i]); 
		}
	}
	for(int i = 0; i <= N; ++i) for(int j = 0; j <= N; ++j) dp[i][j] = inf;
	dp[0][0] = 0;
	for(int i = 0; i <= N; ++i) {
		for(int j = 0; j <= N; ++j) {
			if(i == N && j == N) continue;
			if(i < N) dp[i + 1][j] = min(dp[i + 1][j], dp[i][j] + cost[0][i + 1][j]);
			if(j < N) dp[i][j + 1] = min(dp[i][j + 1], dp[i][j] + cost[1][j + 1][i]);
		}
	}
	cout << dp[N][N] << endl;
}