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.ArrayList;
import java.util.Scanner;

public class Binary_Subsequence_Rotation {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt(); scan.nextLine();
        String s = scan.nextLine(), t = scan.nextLine();
        ArrayList<Integer> a1 = new ArrayList<>(), a2 = new ArrayList<>();
        int sum = 0;
        for (int i = 0; i < n; i++) {
            if (s.charAt(i) != t.charAt(i)) {
                if (s.charAt(i) == '1') {
                    sum++; a1.add(1); a2.add(-1);
                } else {
                    sum--; a1.add(-1); a2.add(1);
                }
            }
        }
        if (sum != 0) {
            System.out.println(-1);
        } else {
            System.out.println(Math.max(kadaneAlgo(a1), kadaneAlgo(a2)));
        }
        
    }
    static int kadaneAlgo(ArrayList<Integer> arr) {
        if (arr.size() == 0) return 0;
        int mxh, msf;
        mxh = msf = arr.get(0);
        for (int i = 1; i < arr.size(); i++) {
            mxh = Math.max(arr.get(i), arr.get(i) + mxh);
            msf = Math.max(msf, mxh);
        }
        return msf;
    }
}
