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 Solution {
	static Scanner sc=new Scanner(System.in);
	static PrintWriter out=new PrintWriter(System.out);
	//debug
	static void debug(Object... o) {
        System.out.println(Arrays.deepToString(o));
    }
	//Main
	public static void main(String args[]) {
		int test=1;
		//test=sc.nextInt();
		while(test-->0) {
			//Think before code, it won't hurt
			int n=sc.nextInt();
			sc.nextLine();
			char s[]=sc.nextLine().toCharArray(),t[]=sc.nextLine().toCharArray();
			ArrayList<Integer> a=new ArrayList<>();
			int z1=0,z2=0;
			for(int i=0;i<n;i++) {
				if(s[i]!=t[i]) a.add(s[i]-'0');
				if(s[i]=='0') z1++;
				if(t[i]=='0') z2++;
			}
			if(z1!=z2) {
				out.println(-1);
				continue;
			}
			int N=a.size(),x=0,y=0,ans1=0,ans2=0;
			//Problem reduced to calculate sum of maximum depth of brackets [] and () like ()((()))[[[]]]
			Stack<Integer> s1=new Stack<>(),s2=new Stack<>();
			for(int i=0;i<N;i++) {
				if(a.get(i)==0) {
					if(!s1.isEmpty()) {
						x--;
						s1.pop();
					}
					else {
						s2.push(1);
						y++;
					}
				}
				else {
					if(!s2.isEmpty()) {
						y--;
						s2.pop();
					}
					else {
						s1.push(1);
						x++;
					}
				}
				ans1=Math.max(ans1, x);
				ans2=Math.max(ans2, y);
			}
			out.println(ans1+ans2);
		}
		out.flush();
		out.close();
	}
}
