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.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeSet;
public class Main {
	static int func(ArrayList<Integer> l) {
		int ans=0;
		int zero=0;
		int one=0;
		int n=l.size();
		for (int i=0;i<n-1;i++) {
			if(l.get(i)==1) {
				if(zero==0) {
					one++;
					ans++;
				}
				else {
					zero--;
					one++;
				}
			}
			else {
				if(one==0) {
					zero++;
					ans++;
				}
				else {
					one--;
					zero++;
				}
			}
		}
		return ans;
	}
	public static void main(String[] args) throws IOException 
	{ 
		FastScanner f = new FastScanner(); 
		int t=1;
//		t=f.nextInt();
		PrintWriter out=new PrintWriter(System.out);
		for(int tt=0;tt<t;tt++) {
			int n=f.nextInt();
			char[] a=f.next().toCharArray();
			char[] b=f.next().toCharArray();
			int count=0;
			int numonea=0;
			int numoneb=0;
			ArrayList<Integer> main=new ArrayList<>();
			
			for(int i=0;i<n;i++) {
				if(a[i]=='1') {
					numonea++;
				}
				if(b[i]=='1') {
					numoneb++;
				}
				if(a[i]!=b[i]) {
					main.add((a[i]=='1')?1:-1);
				}
			}
			if(numonea!=numoneb) {
				System.out.println(-1);
			}
			else {
				System.out.println(func(main));
			}
		}
		out.close();
	} 
	static void sort(int [] a) {
        ArrayList<Integer> q = new ArrayList<>();
        for (int i: a) q.add(i);
        Collections.sort(q);
        for (int i = 0; i < a.length; i++) a[i] = q.get(i);
    }
    
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
		long[] readLongArray(int n) {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
	}
} 	