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
// ※※※ 解答不能 ※※※
// 解き直し.
// sugim48氏.
// https://atcoder.jp/contests/arc097/submissions/2498127
// 正答者の解答を勉強する.
#include <bits/stdc++.h>
using namespace std;
#define rep(i, N) for (int i = 0; i < N; i++)
#define pb push_back
typedef long long ll;

vector<vector<int>> f(vector<int> a, vector<int> b){
    int N = a.size();
    vector<vector<int>> c(N, vector<int>(N + 1));
    rep(x, N) rep(y, N) c[x][y + 1] = c[x][y] + (a[x] < b[y]);
    return c;
}

int main(){
    int N;
    scanf("%d", &N);
    vector<int> a(N), b(N);
    rep(t, N * 2) {
        char s[2];
        scanf("%s", s);
        int i;
        scanf("%d", &i);
        i--;
        if (s[0] == 'W') a[i] = t;
        if (s[0] == 'B') b[i] = t;
    }
    vector<vector<int>> aa = f(a, a), ab = f(a, b), ba = f(b, a), bb = f(b, b);
    vector<vector<int>> dp(N + 1, vector<int>(N + 1, INT_MAX));
    dp[0][0] = 0;
    rep(x, N + 1) rep(y, N + 1) {
        if (x < N) dp[x + 1][y] = min(dp[x + 1][y], dp[x][y] + aa[x][x] + ab[x][y]);
        if (y < N) dp[x][y + 1] = min(dp[x][y + 1], dp[x][y] + bb[y][y] + ba[y][x]);
    }
    printf("%d\n", dp[N][N]);
    return 0;
}