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.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.util.TreeSet;

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

/**
 *
 * @author abacaba
 */
public class e {
    public static void main(String[] args) {
        FS in = new FS(System.in);
        PrintWriter out = new PrintWriter(System.out);
        
//        int t = in.nextInt(); int c = 0; while(c++<t) {
//            out.print("Case #" + c + ": ");
            new e().solve(in, out);
//        }
        
        out.close();
    }
    
    void solve(FS in, PrintWriter out) {
        int n = in.nextInt();
        char[] ss = in.next().toCharArray();
        char[] tt = in.next().toCharArray();
        
        int[] s = new int[n];
        int[] t = new int[n];
        for(int i = 0; i < n; i++) s[i] = ss[i]-'0';
        for(int i = 0; i < n; i++) t[i] = tt[i]-'0';
        
        
        int tot = 0;
        for(int c : s) tot += c;
        for(int c : t) tot -= c;
        if(tot != 0) {
            out.println(-1);
            return;
        }
        
        TreeSet<Integer>[] sets = new TreeSet[] { new TreeSet<>(), new TreeSet<>() };
        
        for(int i = 0; i < n; i++) if(s[i] != t[i]) sets[s[i]].add(i);
        
        sets[0].add(n);
        sets[1].add(n);
        
        int res = 0;
        
        while(sets[0].size() != 1 || sets[1].size() != 1) {
            int curr = Math.min(sets[0].first(), sets[1].first());
            int start = curr;
            sets[s[curr]].pollFirst();
            while(true) {
                int next = sets[s[curr]^1].higher(curr);
                if(next == n) break;
                curr = next;
                sets[s[curr]].remove(curr);
            }
            if(s[curr] == s[start]) sets[s[curr]].add(curr);
            res++;
        }
        out.println(res);
    }
    
    static class FS {
        BufferedReader in; StringTokenizer tk;
        public FS(InputStream st) {
            in = new BufferedReader(new InputStreamReader(st));
            tk = new StringTokenizer("");
        }
        
        public String next() {
            if(tk.hasMoreTokens()) return tk.nextToken();
            try { tk = new StringTokenizer(in.readLine()); } catch(Exception e) {}
            return next();
        }
        
        public int nextInt() { return Integer.parseInt(next()); }
    }
}
