Naman has two binary strings s and t of length n (a binary string is a string which only consists of the characters "0" and "1"). He wants to convert s into t using the following operation as few times as possible.

In one operation, he can choose any subsequence of s and rotate it clockwise once.

For example, if s = 1110100, he can choose a subsequence corresponding to indices (1-based) \{2, 6, 7 \} and rotate them clockwise. The resulting string would then be s = 1010110.

A string a is said to be a subsequence of string b if a can be obtained from b by deleting some characters without changing the ordering of the remaining characters.

To perform a clockwise rotation on a sequence c of size k is to perform an operation which sets c_1:=c_k, c_2:=c_1, c_3:=c_2, …, c_k:=c_{k-1} simultaneously.

Determine the minimum number of operations Naman has to perform to convert s into t or say that it is impossible. 

Input

The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of the strings.

The second line contains the binary string s of length n.

The third line contains the binary string t of length n.

Output

If it is impossible to convert s to t after any number of operations, print -1.

Otherwise, print the minimum number of operations required.

Examples

Input


6
010000
000001


Output


1

Input


10
1111100000
0000011111


Output


5

Input


8
10101010
01010101


Output


1

Input


10
1111100000
1111100001


Output


-1

Note

In the first test, Naman can choose the subsequence corresponding to indices \{2, 6\} and rotate it once to convert s into t.

In the second test, he can rotate the subsequence corresponding to all indices 5 times. It can be proved, that it is the minimum required number of operations.

In the last test, it is impossible to convert s into t.
/*import java.util.*;

public class CodeForces1370D{
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		int n = input.nextInt();
		String s = input.next();
		String t = input.next();
		int count1_one = 0;
		int count2_one = 0;
		for(int i = 0;i<n;i++){
			if(s.charAt(i) == '1'){
				count1_one++;
			}
			if(t.charAt(i) == '1'){
				count2_one++;
			}
		}

		if(count1_one != count2_one){
			System.out.println(-1);
		}
		else{
			int j = 0;
			int count = 0;
			for(int i = 0;i<n;i++){
				if(s.charAt(i) != t.charAt(i)){
					if(s.charAt(i) == '1'){
						count++;
					}
				}
				j++;
			}
			if(count  == count1_one){
				System.out.println(count);
			}
			else{ 
				System.out.println(1 + count);
			}
		}
	}
}*/

import java.io.*;
import java.util.*;
public class CodeForces1370D
{
    public static void main(String[] args) throws IOException
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
 
        int i,N;
 
        N=Integer.parseInt(br.readLine().trim());
        char s[]=br.readLine().trim().toCharArray();
        char t[]=br.readLine().trim().toCharArray();
        
        int sc1=0,tc1=0;
        for(i=0;i<N;i++)
        {
            if(s[i]=='1') sc1++;
            if(t[i]=='1') tc1++;
        }
 
        if(sc1!=tc1)
        {
            System.out.println(-1);
            System.exit(0);
        }
 
        int x0=0,x1=0;
        for(i=0;i<N;i++)
        {
            if(s[i]==t[i]) continue;
            if(s[i]=='0')
            {
                if(x0>0) x0--;
                x1++;
            }
            else
            {
                if(x1>0) x1--;
                x0++;
            }
        }
        System.out.println(x0+x1);
    }
}