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
//IOI 2021

#include <bits/stdc++.h>
using namespace std;

#define pb push_back
#define ff first
#define ss second

typedef long long ll;
typedef pair<int, int> pii;

const int MAXN = 2 * 1000 + 17, inf = 2e9 + 17;

int n, a[2 * MAXN], b[MAXN], w[MAXN], dp[MAXN][MAXN], TB[MAXN], TW[MAXN], PB[MAXN][MAXN], PW[MAXN][MAXN];
char col[2 * MAXN];

int main() {
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n;
	for (int i = 1; i <= 2 * n; i++) {
		cin >> col[i] >> a[i];
		if (col[i] == 'W')
			w[a[i]] = i;
		else
			b[a[i]] = i;
	}
	for (int i = 1; i <= n; i++)
		for (int j = i + 1; j <= n; j++)
			TW[i] += (w[j] < w[i]), TB[i] += (b[j] < b[i]);
	for (int i = 1; i <= n; i++)
		for (int j = n - 1; j >= 0; j--)
			PW[i][j] = PW[i][j + 1] + (b[j + 1] < w[i]), PB[i][j] = PB[i][j + 1] + (w[j + 1] < b[i]);

	for (int i = 0; i <= n; i++)
		for (int j = 0; j <= n; j++)
			dp[i][j] = inf;
	dp[n][n] = 0;
	for (int i = n; i >= 0; i--)
		for (int j = n; j >= 0; j--) {
			if (i != n)
				dp[i][j] = min(dp[i][j], dp[i + 1][j] + TW[i + 1] + PW[i + 1][j]);
			if (j != n)
				dp[i][j] = min(dp[i][j], dp[i][j + 1] + TB[j + 1] + PB[j + 1][i]);
		}
	cout << dp[0][0] << endl;
//	cerr << "# "  << dp[1][0] << endl;
	return 0;
}

