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.*;
import java.io.*;
import java.lang.*;
import java.math.*;

public class E {
    public static void main(String[] args) throws Exception {
        BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        StringTokenizer st = new StringTokenizer(bf.readLine());
        int n = Integer.parseInt(st.nextToken());
        String s = bf.readLine();
        String t = bf.readLine();
        int s1 = 0;
        int t1 = 0;

        int[] a = new int[n];

        for (int i = 0; i < n; i++) {
            if (s.charAt(i) == '1') {
                s1++;
            }
            if (t.charAt(i) == '1') {
                t1++;
            }
            if (s.charAt(i) == t.charAt(i)) {
                a[i] = 0;
            } else {
                if (s.charAt(i) == '1') {
                    a[i] = 1;
                } else {
                    a[i] = - 1;
                }
            }
        }
        if (s1 != t1) {
            out.println(-1);
        } else {
            int maxPos = 0;
            int maxNeg = 0;
            int curPos = 0;
            int curNeg = 0;
            for (int i = 0; i < n; i++) {
                if (a[i] == 1) {
                    if (curPos < 0) curPos = 0;
                    curPos++;
                    curNeg--;
                } else if (a[i] == -1) {
                    if (curNeg < 0) curNeg = 0;
                    curPos--;
                    curNeg++;
                }
                maxPos = max(maxPos, curPos);
                maxNeg = max(maxNeg, curNeg);
            }
            out.println(max(maxPos, maxNeg));
        }

        out.close();
        System.exit(0);
    }

    public static int max(int a, int b) {
        return a > b ? a : b;
    }

    public static int abs(int a) {
        return a < 0 ? -a : a;
    }
}