You are given a bracket sequence s consisting of n opening '(' and closing ')' brackets.

A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.

You can change the type of some bracket s_i. It means that if s_i =  ')' then you can change it to '(' and vice versa.

Your task is to calculate the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.

Input

The first line of the input contains one integer n (1 ≤ n ≤ 10^6) — the length of the bracket sequence.

The second line of the input contains the string s consisting of n opening '(' and closing ')' brackets.

Output

Print one integer — the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.

Examples

Input


6
(((())


Output


3


Input


6
()()()


Output


0


Input


1
)


Output


0


Input


8
)))(((((


Output


0

import java.util.*;
import java.math.*;

public class Main {
	public static void main(String[] args) {
		int N=1000005;
		Scanner cin=new Scanner(System.in);
		int []a=new int[N];
		int []b=new int[N];
		int n;
		String str;
		n=cin.nextInt();
		str=cin.next();
		int cnt=0;
		for(int i=0;i<n;i++) {
			a[i]=cnt;
			if(str.charAt(i)=='(') 
				cnt++;
			else
				cnt--;
			if(i>0&&a[i-1]<0)
				a[i]=-1;
		}
		cnt=0;
		for(int i=n-1;i>=0;i--) {
			b[i]=cnt;
			if(str.charAt(i)==')') 
				cnt++;
			else
				cnt--;
			if(i<n-1&&b[i+1]<0)
				b[i]=-1;
		}
		int ans=0;
		for(int i=0;i<n;i++) {
			if(a[i]<0&&b[i]<0) {
				ans=0;
				break;
			}
			if(a[i]<0||b[i]<0)
				continue;
			int x=a[i]-b[i];
			if(x==1&&str.charAt(i)=='(')
				ans++;
			else if(x==-1&&str.charAt(i)==')')
				ans++;
		}
		System.out.println(ans);
		cin.close();
	}
}
