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.Scanner;
import java.util.TreeSet;

public class binsubrot {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        char[] s = in.next().toCharArray();
        char[] t = in.next().toCharArray();
        TreeSet<Integer> plus = new TreeSet<>(), minus = new TreeSet<>();

        for(int i=0; i<n; i++){
            if(s[i]==t[i]) continue;
            if(s[i]=='0') plus.add(i);
            else minus.add(i);
        }

        if(plus.size() != minus.size()){
            System.out.println(-1);
            return;
        }

        int c = 0;

        while(!plus.isEmpty()){
            TreeSet<Integer> a, b;
            if(plus.first() < minus.first()){
                a = plus;
                b = minus;
            }
            else{
                b = plus;
                a = minus;
            }
            int pos = -1;
            while(true){
                Integer higher = a.higher(pos);
                if(higher==null) break;
                Integer bhigher = b.higher(higher);
                if(bhigher==null) break;
                a.remove(higher);
                b.remove(bhigher);
                pos = bhigher;
            }
            c++;
        }

        System.out.println(c);
    }
}
