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 file {

    private static final Scanner sc = new Scanner(System.in);

    private static int maxi(int x, int[] arr) {
        int curr = 0, max = 0;
        for (int i = 0; i < arr.length; i++) {
            curr += x*arr[i];
            max = Math.max(max, curr);
            if (curr < 0)
                curr = 0;
        }
        return max;
    }


    private static void foo() {
        int n = sc.nextInt(), count1 = 0, count2 = 0;
        String s = sc.next();
        String t = sc.next();
        for (char x : s.toCharArray()) {
            if (x == '1') count1++;
        }
        for (char x : t.toCharArray()) {
            if (x == '1') count2++;
        }
        if (count1 != count2) {
            System.out.println(-1);
            return;
        }
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            if (s.charAt(i) == t.charAt(i))
                arr[i] = 0;
            else if (s.charAt(i) == '1') arr[i] = 1;
            else arr[i] = -1;
        }
        System.out.println(Math.max(maxi(1, arr), maxi(-1, arr)));
    }


    public static void main(String[] args) {
        //int t = sc.nextInt();
        //for (int i = 0; i < t; i++) {
        //    foo();
        //}
        foo();
    }
}