CodeForces-723B(记录一下自己的失误。。。)

B. Text Document Analysis

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Modern text editors usually show some information regarding the document being edited. For example, the number of words, the number of pages, or the number of characters.   In this problem you should implement the similar functionality.

You are given a string which only consists of:

• uppercase and lowercase English letters,
• underscore symbols (they are used as separators),
• parentheses (both opening and closing).

It is guaranteed that each opening parenthesis has a succeeding closing parenthesis. Similarly, each closing parentheses has a preceding opening parentheses matching it. For each pair of matching parentheses there are no other parenthesis between them. In other words, each parenthesis in the string belongs to a matching "opening-closing" pair, and such pairs can't be nested.

For example, the following string is valid: "_Hello_Vasya(and_Petya)__bye_(and_OK)".

Word is a maximal sequence of consecutive letters, i.e. such sequence that the first character to the left and the first character to the right of it is an underscore, a parenthesis, or it just does not exist. For example, the string above consists of seven words: "Hello", "Vasya", "and", "Petya", "bye", "and" and "OK". Write a program that finds:

• the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
• the number of words inside the parentheses (print 0, if there is no word inside the parentheses).

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 255) — the length of the given string. The second line contains the string consisting of only lowercase and uppercase English letters, parentheses and underscore symbols.

Output

Print two space-separated integers:

• the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
• the number of words inside the parentheses (print 0, if there is no word inside the parentheses).

Examples

Input

37
_Hello_Vasya(and_Petya)__bye_(and_OK)

Output

5 4

Input

37
_a_(_b___c)__de_f(g_)__h__i(j_k_l)m__

Output

2 6

Input

27
(LoooonG)__shOrt__(LoooonG)

Output

5 2

Input

5
(___)

Output

0 0

Note

In the first sample, the words "Hello", "Vasya" and "bye" are outside any of the parentheses, and the words "and", "Petya", "and" and "OK" are inside. Note, that the word "and" is given twice and you should count it twice in the answer.

分析：

         所有的word被’_‘ ,  ’( ‘ 或  ’)‘分割，统计括号内的word个数，括号外最长的word长度，不存在括号嵌套。然而笔者却犯了个很低级的错误，访问越界了。。。结果WA两次。警示自己以后写出单变量双重循环时，一定要在内部循环加上越界条件。。。。

AC代码：

#include<iostream>

#include<cstring>

#include<math.h>

#include<stdlib.h>

#include<cstring>

#include<cstdio>

#include<algorithm>

using namespace std;

const int Max = 1e5+5;

const int mod = 1e9+7;

char arr[Max];

int n;

int main( )

{

    //freopen("input.txt","r",stdin);

    while(scanf("%d", &n)!=EOF)

    {

        cin>>arr;

        int counter = 0, MAX = 0;

        int flag = 1;//表示在括号外

        for(int i=0; i<n; i++)

        {

            if(arr[i] == '(')

            {

                flag = 0;

                continue;

            }

            if(arr[i] == ')')

            {

                flag = 1;

                continue;

            }

            if(flag == 1)

            {

                int a = 0;

                //这个i<k坑了的WA了两次。。。

                while(arr[i]!='(' && arr[i]!=')' && arr[i]!='_' && i<n)

                {

                    a++;

                    i++;

                }

                if(a>0)

                    i--;

                MAX = max(MAX, a);

            }

            if(flag == 0)

            {

                int a = 0;

                while(arr[i]!='(' && arr[i]!=')' && arr[i]!='_')

                {

                    a++;

                    i++;

                }

                if(a > 0)

                {

                    counter++;

                    i--;

                }

            }

        }

        printf("%d %d\n", MAX, counter);

    }

    return 0;

}