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 ≤ 10^6) — the length of Dima's sequence.

The second line contains string of length n, 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.
import java.util.*;
import java.io.*;
import java.math.*;

public class Main {

	//static int p = 1000000007;
	static int p = 998244353;
	
	
	

	public static void main(String args[]) throws IOException {
		Scanner sc = new Scanner();
		PrintWriter out = new PrintWriter(System.out);
		//Mod mod = new Mod();
		//mod.precalc();

		int t = 1;
		//t = sc.nextInt();

		
		while (t-- > 0) {
			
			int n = sc.nextInt(), arr1[]=new int[n+1];
			char arr[]=sc.next().toCharArray();
			for(int i=0; i<n; i++) {
				if(arr[i]==')')
					arr1[i+1]=arr1[i]-1;
				else
					arr1[i+1]=arr1[i]+1;
			}
			
			int prev = arr1[0], ans=0;
			for(int i=1; i<=n; i++) {
				if(arr1[i]==0) {
					if(arr1[i-1]<0)
						ans+=(i-prev);
					prev=i;
				}
			}
			
			if(arr1[n]!=0)
				out.print(-1);
			else
				out.print(ans);
			
		}
		out.flush();
	}
	
	
	
    
	static int count(int x, int[] arr) {
		int count = 0;

		for (int i = 0; i < arr.length; i++)
			if (arr[i] == x)
				count++;
		return count;
	}

	public static String reverseString(String str) {
		StringBuilder sb = new StringBuilder(str);
		sb.reverse();
		return sb.toString();
	}

	static boolean isPalindrome(String str) {
		int i = 0, j = str.length() - 1;

		while (i < j) {

			if (str.charAt(i) != str.charAt(j))
				return false;

			i++;
			j--;
		}
		return true;
	}

	static void print(int[] arr) {
		for (int i = 0; i < arr.length; i++)
			System.out.print(arr[i] + " ");
		System.out.println();
	}

	public static int abs(int x) {
		return ((x > 0) ? x : -x);
	}

	public static long abs(long x) {
		return ((x > 0) ? x : -x);
	}

	public static int max(int a, int b) {
		return Math.max(a, b);
	}

	public static long max(long a, long b) {
		return Math.max(a, b);
	}

	public static int min(int a, int b) {
		return Math.min(a, b);
	}

	public static long min(long a, long b) {
		return Math.min(a, b);
	}

	static int gcd(int a, int b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}

	static int modInverse(int a, int m) {
		int g = gcd(a, m);
		if (g != 1)
			return -1;
		else
			return power(a, m - 2, m);

	}

	// To compute x^y under modulo m
	static int power(int x, int y, int m) {
		if (y == 0)
			return 1;

		int p = power(x, y / 2, m) % m;
		p = (p * p) % m;

		if (y % 2 == 0)
			return p;
		else
			return (x * p) % m;
	}

	static int[] primeGenerator(int num) {
		int length = 0, arr[] = new int[num], a = num, factor = 1;
		if (num % 2 == 0) {
			while (num % 2 == 0) {
				num /= 2;
				factor *= 2;
			}
			arr[length++] = factor;
		}
		for (int i = 3; i * i <= a; i++) {
			factor = 1;
			if (num % i == 0) {
				while (num % i == 0) {
					num /= i;
					factor *= i;
				}
				arr[length++] = factor;
			}
		}
		if (num > 1)
			arr[length++] = num;
		return Arrays.copyOfRange(arr, 0, length);
	}

	static boolean isPrime(int n) {
		// Corner cases
		if (n <= 1)
			return false;
		if (n <= 3)
			return true;

		// This is checked so that we can skip
		// middle five numbers in below loop
		if (n % 2 == 0 || n % 3 == 0)
			return false;

		for (int i = 5; i * i <= n; i = i + 6)
			if (n % i == 0 || n % (i + 2) == 0)
				return false;

		return true;
	}

	
	  static class Scanner { BufferedReader br; StringTokenizer st;
	  
	 Scanner() { br = new BufferedReader(new InputStreamReader(System.in)); }
	  
	  Scanner(String fileName) throws FileNotFoundException { br = new
	  BufferedReader(new FileReader(fileName)); }
	  
	  String next() throws IOException { while (st == null || !st.hasMoreTokens())
	  st = new StringTokenizer(br.readLine()); return st.nextToken(); }
	  
	  String nextLine() throws IOException { return br.readLine(); }
	  
	  int nextInt() throws IOException { return Integer.parseInt(next()); }
	  
	  long nextLong() throws NumberFormatException, IOException { return
	  Long.parseLong(next()); }
	  
	  double nextDouble() throws NumberFormatException, IOException { return
	  Double.parseDouble(next()); }
	  
	  boolean ready() throws IOException { return br.ready(); }
	  
	  }
	 

	static void sort(int[] a) {
		shuffle(a);
		Arrays.sort(a);
	}

	static void shuffle(int[] a) {
		int n = a.length;
		Random rand = new Random();
		for (int i = 0; i < n; i++) {
			int tmpIdx = rand.nextInt(n);
			int tmp = a[i];
			a[i] = a[tmpIdx];
			a[tmpIdx] = tmp;
		}
	}
}

class Pair {
	int x;
	int y;

	Pair(int a, int b) {
		x = a;
		y = b;
	}

	void print() {
		System.out.println(this.x + " " + this.y);
	}
}

class Mod{
	static final int MOD = 998244353;
	
	int add(int x, int y){
		x += y;
	    while(x >= MOD) x -= MOD;
	    while(x < 0) x += MOD;
	    return x;
	}
	
	int subtract(int x, int y) {
		x -= y;
	    while(x >= MOD) x -= MOD;
	    while(x < 0) x += MOD;
	    return x;
	}

	int mul(int x, int y)
	{
		int result = (int)((long)x*y % MOD);
	    return result>=0?result:result+MOD;
	}

	int binpow(int x, int y)
	{
	    int z = 1;
	    while(y!=0)
	    {
	        if((y & 1)!=0) 
	        	z = mul(z, x);
	        x = mul(x, x);
	        y >>= 1;
	    }
	    return z;
	}

	int inv(int x)
	{
	    return binpow(x, MOD - 2);
	}

	int divide(int x, int y)
	{
	    return mul(x, inv(y));
	}
	
	static final int N =200043;
	int[] fact;
	
	Mod() {
		fact = new int[N];
	}
	
	void precalc()
	{
	    fact[0] = 1;
	    for(int i = 1; i < N; i++)
	        fact[i] = mul(fact[i - 1], i);
	}

	int C(int n, int k)
	{
	    return divide(fact[n], mul(fact[k], fact[n - k]));
	}
}
