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 <vector>

using namespace std;

int main(){
	int N;
	while(cin >> N){
		char c; int a;
		vector<int> v;
		for(int i=0;i<2*N;i++){
			cin >> c >> a; --a;
			v.push_back(c == 'W' ? a : a + N);
		}
		vector<vector<int>> cntW(N, vector<int>(N+1, 0));
		vector<vector<int>> cntB(N, vector<int>(N+1, 0));
		vector<int> flipW(N, 0);
		vector<int> flipB(N, 0);
		for(int i=0;i<2*N;i++){
			if(v[i] < N){
				for(int j=0;j<i;j++){
					if(v[j] < N){
						if(v[j] > v[i]) ++flipW[v[i]];
					} else {
						++cntW[v[i]][v[j]-N];
					}
				}
				for(int j=N-2;j>=0;j--) cntW[v[i]][j] += cntW[v[i]][j+1];
			} else {
				for(int j=0;j<i;j++){
					if(v[j] < N){
						++cntB[v[i]-N][v[j]];
					} else {
						if(v[j] > v[i]) ++flipB[v[i]-N];
					}
				}
				for(int j=N-2;j>=0;j--) cntB[v[i]-N][j] += cntB[v[i]-N][j+1];
			}
		}
		vector<vector<int>> dp(N+1, vector<int>(N+1, 4*N*N));
		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] + flipW[i] + cntW[i][j]);
				}
				if(j < N){
					dp[i][j+1] = min(dp[i][j+1], dp[i][j] + flipB[j] + cntB[j][i]);
				}
			}
		}
		cout << dp[N][N] << endl;
	}
}