Okabe and Boxes(栈)

题目：

Okabe and Super Hacker Daru are stacking and removing boxes. There are n boxes numbered from 1 to n. Initially there are no boxes on the stack.

Okabe, being a control freak, gives Daru 2n commands: n of which are to add a box to the top of the stack, and n of which are to remove a box from the top of the stack and throw it in the trash. Okabe wants Daru to throw away the boxes in the order from 1 to n. Of course, this means that it might be impossible for Daru to perform some of Okabe's remove commands, because the required box is not on the top of the stack.

That's why Daru can decide to wait until Okabe looks away and then reorder the boxes in the stack in any way he wants. He can do it at any point of time between Okabe's commands, but he can't add or remove boxes while he does it.

Tell Daru the minimum number of times he needs to reorder the boxes so that he can successfully complete all of Okabe's commands. It is guaranteed that every box is added before it is required to be removed.

Input

The first line of input contains the integer n (1 ≤ n ≤ 3·105) — the number of boxes.

Each of the next 2n lines of input starts with a string "add" or "remove". If the line starts with the "add", an integer x (1 ≤ x ≤ n) follows, indicating that Daru should add the box with number x to the top of the stack.

It is guaranteed that exactly n lines contain "add" operations, all the boxes added are distinct, and n lines contain "remove" operations. It is also guaranteed that a box is always added before it is required to be removed.

Output

Print the minimum number of times Daru needs to reorder the boxes to successfully complete all of Okabe's commands.

Examples

Input

3
add 1
remove
add 2
add 3
remove
remove

Output

1

Input

7
add 3
add 2
add 1
remove
add 4
remove
remove
remove
add 6
add 7
add 5
remove
remove
remove

Output

2

Note

In the first sample, Daru should reorder the boxes after adding box 3 to the stack.

In the second sample, Daru should reorder the boxes after adding box 4 and box 7 to the stack.

需要一个栈来处理，如果需要调整的就将栈清空，因为如果你调整之后，状态肯定是最优的，最大值在下边，最小在最上边。如果你插入的值就是你想要的结果，那么就将其弹出。如果栈空，什么事都不用干。

#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
const int maxn = 3e5+5;
int a[maxn];//栈
int u[maxn];
char s[10];
int main()
{
    int n;
    int l=0;
    int k;
    int tail=-1;
    cin>>n;
    n=n*2;
    int i=1;
    while(n--)
    {
        scanf("%s",s);
        if(s[0]=='a')
            {
                cin>>k;
                tail++;
                a[tail]=k;
               // cout<<a[tail]<<endl;
            }
        else if(s[0]=='r')
        {
              if(tail==0);
              else if(a[tail]==i)tail--;//目标值就将他弹出；
              else l++,tail=0;//需要调整的时候就将他清空。因为每一次调整过后都会是最优，也就是说都会是大的在下，小的在上。
                                    //即便下边有最大的。上边只保留新插入的值。
              i++;
        }

    }
    cout<<l<<endl;
    return 0;
}