Codeforces 723B Text Document Analysis （字符串处理）

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.

思路：

题意就是给你一串字符串，查询在括号外边的所有字符中能构成的单词中的最大长度（中间有‘_’的单词就被分开了）和括号里边能构成的单词的个数。只需要对“_”、“(”、")"进行分情况讨论即可。需要注意的两个WA点是，如果出现"_"的时候前面是没有单词的，那么它之前的数就不能构成一个单词了；在一个就是当判断字符串的最后一个字符时候要在它continue之前完成maxn的更新。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		char s[256];
		scanf("%s",s);
		int maxn1=-1;
		int maxn2=0;
		int cnt1=0;
		int cnt2=0;
		int flag=false;
		for(int i=0;i<n;i++)
		{
			if(flag==false&&s[i]!='_'&&s[i]!='('&&s[i]!=')')
			{
				cnt1++;
				if(i==n-1&&cnt1!=0)//WA在这个细节了 
				{
					maxn1=max(maxn1,cnt1);
				}
				continue; 
			} 
			if(flag==false&&s[i]=='_')
			{
				if(cnt1==0)//WA在这个细节了 
				{
	     			cnt1=0;
					continue;	
				}
				else
				{		
					maxn1=max(maxn1,cnt1);
					cnt1=0;
					continue;	
				}
			} 
			if(flag==false&&s[i]=='(')
			{
				if(cnt1==0)
				{
					cnt1=0;
					flag=true;
					continue;
				}
				else
				{
					maxn1=max(maxn1,cnt1);
					cnt1=0;
					flag=true;
					continue;					
				}
			} 
			if(flag==true&&s[i]!='_'&&s[i]!='('&&s[i]!=')')
			{
				cnt2++;
				continue;
			}
			if(flag==true&&s[i]=='_')
			{
				if(cnt2!=0)maxn2++;
				cnt2=0;
				continue;
			}
			if(flag==true&&s[i]==')')
			{
				flag=false;
		    	if(cnt2!=0)maxn2++;
				cnt2=0;
				continue;
			}
		}
		if(maxn1==-1)printf("0 %d\n",maxn2);
		else printf("%d %d\n",maxn1,maxn2);
	}
	return 0;
}