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.*;
import java.util.*;
 

public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Unstoppable solver = new Unstoppable();
    //  int t=in.nextInt();
	 // while(t-->0)
        solver.solve(in, out);
        out.close();
    }
 
    static class Unstoppable { static long n,k,ans;

       
     public void solve(InputReader in, PrintWriter out) {
       int  n=in.nextInt();
      String s=in.next();
      String t=in.next();
      int one=0,zero=0;
      int dp[][]=new int[2][2];
      ArrayList<Integer> array=new ArrayList<>();
      for(int i=0;i<n;i++){
          if(s.charAt(i)!=t.charAt(i)){  if(t.charAt(i)=='1') { array.add(1); one++;} else { array.add(0); zero++;}   }
      }
      if(one!=zero) { out.println("-1"); return; }
      for(int i=0;i<array.size();i++){
          if(array.get(i)==0){
              if(dp[1][1]>0) { dp[1][1]--; dp[1][0]++; }
              else if(dp[0][1]>0){ dp[0][1]--; dp[0][0]++; }
              else dp[0][0]++;
           }
          else{
              if(dp[0][0]>0) { dp[0][0]--; dp[0][1]++; }
              else if(dp[1][0]>0){ dp[1][0]--; dp[1][1]++; }
              else dp[1][1]++;
          }
      }
      out.println(dp[0][1]+dp[0][0]+dp[1][0]+dp[1][1]);
    }
}

 
    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
 
        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }
 
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        public long nextLong(){
            return Long.parseLong(next());
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
 
    }
}
