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.*;
public class Main {
    static Scanner in;
    public static void main(String[] args) {
        in = new Scanner(System.in);
        // int num = in.nextInt();
        for (int i = 0; i < 1; i++) {
            int n = in.nextInt(); 
            boolean[] a = new boolean[n];
            int ac = 0;
            String as = in.next();
            String bs = in.next();
            for (int j = 0; j < n; j++) {
                if (a[j] = (as.charAt(j) == '1')) ac++;
            }
            boolean[] b = new boolean[n];
            int bc = 0;
            for (int j = 0; j < n; j++) {
                if (b[j] = (bs.charAt(j) == '1')) bc++;
            }
            if (ac != bc) {
                System.out.println(-1);
                return;
            }
            // memo = new Map[n][k];
            System.out.println(solve(n, a, b));
        }
    }
    
    // 11110000
    // 00001111
    // 
    
    
    // 11001001 -> 100101 -> 1111000011000011
    // 10011100 -> 
    // 
    
    // 10101010
    // 01010101
    // 0 1 2 3 4 5 6 7 
    // 1 0 1 0 1 0 1 0
    
    static int solve(int n, boolean[] a, boolean[] b) {
        List<Boolean> l = new ArrayList<>();
        Boolean f = null;
        int ac = 0;
        int bc = 0;
        int mxa = 0;
        int mxb = 0;
        for (int i = 0; i < n; i++) {
            if (a[i] != b[i]) {
                if (a[i]) {
                    if (bc == 0) ac++;
                    else bc--;
                } else {
                    if (ac == 0) bc++;
                    else ac--;
                }
                mxa = Math.max(mxa, ac);
                mxb = Math.max(mxb, bc);
            }
        }
        return mxa + mxb;
    }
    

    
}