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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.EnumSet;
import java.util.StringTokenizer;

public class Solution {
    static final FS sc = new FS();
    static final PrintWriter pw = new PrintWriter(System.out);

    public static void main(String[] args) {
        int n = sc.nextInt();
        String s = sc.next();
        String t = sc.next();
        int[] diff = new int[n];
        int sum = 0;
        for(int i = 0;i < n;i++) {
            diff[i] = s.charAt(i) - t.charAt(i);
            sum += diff[i];
        }
        if (sum != 0) {
            pw.println(-1);
        } else {
            int cur1 = 0, cur2 = 0, max1 = 0, max2 = 0;
            for(int i = 0;i < n;i++) {
                cur1 += diff[i];
                cur2 -= diff[i];
                max1 = Math.max(max1, cur1);
                max2 = Math.max(max2, cur2);
                cur1 = Math.max(cur1, 0);
                cur2 = Math.max(cur2, 0);
            }
            pw.println(Math.max(max1, max2));
        }
        pw.flush();
    }

    static class FS {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");
        String next() {
            while(!st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch(Exception ignored) {}
            }
            return st.nextToken();
        }
        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
    }
}