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 A {

	static PrintWriter out;
	
	static class FIO {
		BufferedReader in;
		StringTokenizer st = null;
		FIO() throws IOException{
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			st = null;
		}
		FIO(String s) throws IOException {
			in = new BufferedReader(new FileReader(s + ".in"));
			out = new PrintWriter(new FileWriter(s + ".out"));
			st = null;
		}
		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(in.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}
	static FIO in;
	public static void main(String[] args) throws IOException {
		in = new FIO();
		new solver();
		
		out.close();
		
	}
	static class solver {
		
		solver() {
			int N = in.nextInt();
			String s = in.next(), t = in.next();
			int[] a = new int[N], b = new int[N];
			
			int[] oc = new int[2];
			for (int i = 0; i < N; ++i) {
				a[i] = s.charAt(i) - '0';
				b[i] = t.charAt(i) - '0';
				if (a[i] == 0)
					oc[0]++;
				if (b[i] == 0)
					oc[1]++;
			}
			if (oc[0] != oc[1]) {
				out.println(-1);
				return;
			}
			Queue<Integer> q = new LinkedList<>();
			for (int i = 0; i < N; ++i) {
				a[i] -= b[i];
				if (a[i] != 0)
					q.add(a[i]);
			}
			int[] seq = new int[2];
			while (!q.isEmpty()) {
				int x = q.peek();
				q.poll();
				if (x == -1) {
					seq[1] = Math.max(0, seq[1] - 1);
					seq[0]++;
				}
				else {
					seq[0] = Math.max(0, seq[0] - 1);
					seq[1]++;
				}
			}
			out.println(seq[0] + seq[1]);
			
		}
		
	}
	

}
/*





 
*/
