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>
using namespace std;

const int N = 2005;

struct BIT{
    int d[2 * N];
    void i(){ fill(d + 1, d + 2 * N, 0); }
    void u(int x){ for(; x < 2 * N; x += x & -x) d[x]++; }
    void r(int x){ for(; x < 2 * N; x += x & -x) d[x]--; }
    int g(int x){ int r = 0; for(; x; x -= x & -x) r += d[x]; return r; }
} B;

int n, b[N], w[N], d[N][N];

int main(){
    scanf("%d", &n);
    for(int i = 1; i <= 2 * n; i++){
        char s[3]; int x;
        scanf("%s%d", s, &x);
        if(s[0] == 'B') b[x] = i;
        else w[x] = i;
    }
    for(int i = 0; i <= n; i++) fill(d[i], d[i] + n + 1, 4 * N * N);
    d[0][0] = 0;
    for(int i = 0; i <= n; i++){
        B.i();
        for(int k = i + 1; k <= n; k++) B.u(b[k]);
        for(int k = 1; k <= n; k++) B.u(w[k]);
        for(int j = 0; j <= n; j++){
            if(i < n) d[i + 1][j] = min(d[i + 1][j], d[i][j] + B.g(b[i + 1] - 1));
            if(j < n) d[i][j + 1] = min(d[i][j + 1], d[i][j] + B.g(w[j + 1] - 1));
            if(j < n) B.r(w[j + 1]);
        }
    }
    printf("%d\n", min(d[n][n - 1], d[n - 1][n]));
}