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.
// Working program using Reader Class 
import java.io.*; 
import java.util.*; 

public class Main
{ 
	public static void main(String[] args) throws IOException 
	{ 
		FastScanner fs=new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
		int t=1;
		while(t-->0){
			int n=fs.nextInt();
			char c1[]=fs.next().toCharArray();
			char c2[]=fs.next().toCharArray();
			TreeSet<Integer> plus=new TreeSet<Integer>(),minus=new TreeSet<Integer>();
			for(int i=0;i<n;i++){
				if(c1[i]==c2[i])continue;
				if(c1[i]=='1')plus.add(i);
				else minus.add(i);
			}
			if(plus.size()!=minus.size()){
				out.println("-1");
				break;
			}
			int ans=0;
			while(!plus.isEmpty()){
				TreeSet<Integer> a=null,b=null;
				if(plus.first()<minus.first()){
					a=plus;
					b=minus;
				}else{
					a=minus;
					b=plus;
				}
				int pos=-1;
				while(true){
					Integer ahigher=a.higher(pos);
					if(ahigher==null)break;
					Integer bhigher=b.higher(ahigher);
					if(bhigher==null)break;
					a.remove(ahigher);
					b.remove(bhigher);
					pos=bhigher;	
					//System.out.println(ahigher+" "+bhigher+"->"+ans);
				}
				ans++;
				//System.out.println(a.size()+" "+b.size());
			}
			out.println(ans);
		}
		out.flush();
	}
	
}

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[] nextIntArray(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());
	}
	long[] nextLongArray(int n) {
		long[] a=new long[n];
		for (int i=0; i<n; i++) a[i]=nextLong();
		return a;
	}
}