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.*;
public class A{
    static PrintWriter out;
    static InputReader in;
    public static void main(String args[]){
        out = new PrintWriter(System.out);
        in = new InputReader();
        new A();
        out.flush(); out.close();
    }   
    A(){
        solve();
    }
    TreeSet<Integer> ts[] = new TreeSet[2];
    void solve(){
        int n = in.nextInt();
        char s[] = in.next().trim().toCharArray();
        char r[] = in.next().trim().toCharArray();
        for(int i = 0; i < 2; i++)ts[i] = new TreeSet<>();
        for(int i = 0; i < n; i++){
            if(s[i] != r[i]){
                ts[s[i] - '0'].add(i);
            }
        }
        int ans = 0;
        boolean ex = true;
        while(ts[0].size() > 0 && ts[1].size() > 0){
            int t = 0;
            if(ts[0].first() > ts[1].first())t = 1;
            int prv = -1, cnt = 0;
            while(ts[t].size() > 0){
                Object now = ts[t].higher(prv);
                if(now == null)break;
                prv = (int)now;
                ts[t].remove(prv);
                cnt++; t = 1 - t;
            }
            if(cnt < 2){
                ex = false;
                break;
            }else if(cnt % 2 == 1){
                ts[1 - t].add(prv);
            }
            ans++;
        }
        if(ts[0].size() > 0 || ts[1].size() > 0)ex = false;
        out.print(ex ? ans : -1);
    }
    public static class InputReader{
        BufferedReader br;
        StringTokenizer st;
        InputReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        public int nextInt(){
            return Integer.parseInt(next());
        }
        public long nextLong(){
            return Long.parseLong(next());
        }
        public double nextDouble(){
            return Double.parseDouble(next());
        }
        public String next(){
            while(st == null || !st.hasMoreTokens()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch(IOException e){}
            }
            return st.nextToken();
        }
    }
}
        