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 {
	static Scanner sc = new Scanner(System.in);
	static PrintWriter out = new PrintWriter(System.out);
	
	static int solve(TreeSet<Integer> st[]) {
		int cnt = 0;
		while(!st[0].isEmpty()) {
			int cur = st[0].first() < st[1].first() ? 0 : 1;
			int idx = st[cur].first();
			while(true) {
				if(st[cur ^ 1].higher(idx) != null) {
					int nxt = st[cur ^ 1].higher(idx);
					st[cur ^ 1].remove(nxt);
					st[cur].remove(idx);
					
					if(st[cur].higher(nxt) != null) {
						idx = st[cur].higher(nxt);
					}else break;
						
				}else break;
			}
			cnt++;
		}
		return cnt;
	}

	public static void main(String[] args) throws Exception {
		int n = sc.nextInt(), arr[] = new int[n];
		char s[] = sc.nextLine().toCharArray();
		char t[] = sc.nextLine().toCharArray();

		TreeSet<Integer> st[] = new TreeSet[2];
		for (int i = 0; i < 2; i++) {
			st[i] = new TreeSet<>();
		}

		int sum = 0;
		for (int i = 0; i < n; i++) {
			if (s[i] != t[i]) {
				sum += s[i] == '1' ? 1 : -1;
				st[s[i] - '0'].add(i);
			}
		}
		
		out.println(sum == 0 ? solve(st) : - 1);
		out.close();
	}

}

class Scanner {
	StringTokenizer st;
	BufferedReader br;

	public Scanner(InputStream system) {
		br = new BufferedReader(new InputStreamReader(system));
	}

	public Scanner(String file) throws Exception {
		br = new BufferedReader(new FileReader(file));
	}

	public String next() throws IOException {
		while (st == null || !st.hasMoreTokens())
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}

	public String nextLine() throws IOException {
		return br.readLine();
	}

	public int nextInt() throws IOException {
		return Integer.parseInt(next());
	}

	public double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}

	public Long nextLong() throws IOException {
		return Long.parseLong(next());
	}
}