Unusual Competitions

A bracketed sequence is called correct (regular) if by inserting “+” and “1” you can get a well-formed mathematical expression from it. For example, sequences “(())()”, “()” and “(()(()))” are correct, while “)(”, “(()” and “(()))(” are not.

The teacher gave Dmitry’s class a very strange task — she asked every student to come up with a sequence of arbitrary length, consisting only of opening and closing brackets. After that all the students took turns naming the sequences they had invented. When Dima’s turn came, he suddenly realized that all his classmates got the correct bracketed sequence, and whether he got the correct bracketed sequence, he did not know.

Dima suspects now that he simply missed the word “correct” in the task statement, so now he wants to save the situation by modifying his sequence slightly. More precisely, he can the arbitrary number of times (possibly zero) perform the reorder operation.

The reorder operation consists of choosing an arbitrary consecutive subsegment (substring) of the sequence and then reordering all the characters in it in an arbitrary way. Such operation takes l nanoseconds, where l is the length of the subsegment being reordered. It’s easy to see that reorder operation doesn’t change the number of opening and closing brackets. For example for “))((” he can choose the substring “)(” and do reorder “)()(” (this operation will take 2 nanoseconds).

Since Dima will soon have to answer, he wants to make his sequence correct as fast as possible. Help him to do this, or determine that it’s impossible.

Input

The first line contains a single integer n(1≤n≤106)n (1≤n≤10^6)n(1≤n≤106)— the length of Dima’s sequence.

The second line contains string of length nnn, consisting of characters “(” and “)” only.

Output

Print a single integer — the minimum number of nanoseconds to make the sequence correct or “-1” if it is impossible to do so.

Examples
Input
8
))((())(
Output
6
Input
3
(()
Output
-1

Note

In the first example we can firstly reorder the segment from first to the fourth character, replacing it with “()()”, the whole sequence will be “()()())(”. And then reorder the segment from the seventh to eighth character, replacing it with “()”. In the end the sequence will be “()()()()”, while the total time spent is 4+2=6 nanoseconds.

思路

题意：给出长度为n的有左括号和有括号的数组，我们可以每次选择长度为m的连续序列使其满足匹配要求，同时会消耗m秒，求使得整个数组满足匹配要求需要的最小时间。
思路：直接遍历，记录左右括号的数量，当数量相等时，判断是否满足匹配，如果不能，那么这一段序列需要重排，花费该段长度时间~

代码如下

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#define INF 0x3f3f3f3f
typedef long long ll;
using namespace std;
const int mx=1e6+10;
int a,b,n,m,ans;//a记录左括号，b记录右括号,m判断是否匹配
char s[mx];//似乎不需要数组了= =int main()
{a=b=m=0;scanf("%d",&n);ans=0;getchar();for(int i=1;i<=n;i++){scanf("%c",&s[i]);if(s[i]=='('){a++;m++;}else {b++;if(m>0)m--;}if(a==b){if(m!=0)ans+=2*a;a=b=m=0;//一定要记住清零}}if(a!=b)printf("-1\n");elseprintf("%d\n",ans);return 0;
}