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<bits/stdc++.h>
typedef long long ll ;
#define rep(i, a, b) for (int i = a; i <= b; ++i)
using namespace std;

const int MAXN = 2333;
const ll INF = 2e9;
int n;
int dp[MAXN][MAXN];
int a[MAXN + MAXN], c[MAXN + MAXN];
int p1[MAXN], p0[MAXN];
int t[MAXN + MAXN];

void add (int k, int d) { while (k <= n + n) { t[k] += d; k += k & -k; } }
int sum (int k) { int s = 0; while (k > 0) { s += t[k]; k -= k & -k; } return s; }

int main() {
	cin >> n;
	rep(i, 1, n + n) {
		char ch;
		cin >> ch >> a[i];
		if (ch == 'W') {
			p0[a[i]] = i;
			c[i] = 0;
		}
		else {
			p1[a[i]] = i;
			c[i] = 1;
		}
		add(i, 1);
	}
	p1[0] = n + n + 1;
	p0[0] = n + n + 1;
	dp[0][0] = 0;
	rep(j, 1, n) {
		add(p1[j], -1);
		dp[0][j] = dp[0][j - 1] + sum(p1[j] - 1);
	}
	rep(j, 0, n) add(p1[j], 1);
	rep(i, 1, n) {
		add(p0[i], -1);
		rep(j, 0, n) {
			add(p1[j], -1);
			dp[i][j] = INF;
			if (i) dp[i][j] = min(dp[i][j], dp[i - 1][j] + sum(p0[i] - 1));
			if (j) dp[i][j] = min(dp[i][j], dp[i][j - 1] + sum(p1[j] - 1));
		}
		rep(j, 0, n) add(p1[j], 1);
	}
	cout << dp[n][n] << "\n";
	return 0;
} 
