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
{
	
	private void solve()throws Exception
	{
		int n=nextInt();
		String s=nextLine();
		String t=nextLine();
		int cnt=0;
		for(int i=0;i<n;i++)
		{
			if(s.charAt(i)=='1')
				cnt++;
			if(t.charAt(i)=='1')
				cnt--;
		}
		if(cnt!=0)
		{
			out.println(-1);
			return;
		}
		TreeSet<Integer> set[]=new TreeSet[2];
		for(int i=0;i<=1;i++)
			set[i]=new TreeSet<>();
		for(int i=0;i<n;i++)
		{
			if(s.charAt(i)!=t.charAt(i))
				set[s.charAt(i)-'0'].add(i);
		}
		int ans=0;
		while(!set[0].isEmpty())
		{
			Integer curr=-1;
			while(true)
			{
				if(curr==-1)
				{
					Integer v1=set[0].first();
					Integer v2=set[1].first();
					curr=v1<v2?v1:v2;
				}
				else
					curr=set[1-s.charAt(curr)+'0'].higher(curr);
				if(curr==null)
					break;
				Integer next=set[1-s.charAt(curr)+'0'].higher(curr);
				if(next==null)
					break;
				set[s.charAt(curr)-'0'].remove(curr);
				set[s.charAt(next)-'0'].remove(next);
				curr=next;
			}
			ans++;
		}
		out.println(ans);
	}

	 
	///////////////////////////////////////////////////////////

	public void run()throws Exception
	{
		br=new BufferedReader(new InputStreamReader(System.in));
		st=null;
		out=new PrintWriter(System.out);

		solve();
		
		br.close();
		out.close();
	}
	public static void main(String args[])throws Exception{
		new Main().run();
	}
	BufferedReader br;
	StringTokenizer st;
	PrintWriter out;
	String nextToken()throws Exception{
		while(st==null || !st.hasMoreTokens())
		st=new StringTokenizer(br.readLine());
		return st.nextToken();
	}
	String nextLine()throws Exception{
		return br.readLine();
	}
	int nextInt()throws Exception{
		return Integer.parseInt(nextToken());
	}
	long nextLong()throws Exception{
		return Long.parseLong(nextToken());
	}
	double nextDouble()throws Exception{
		return Double.parseDouble(nextToken());
	}
}