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.util.*;
import java.io.*;

public class binaryrot {
    int N;
    boolean[] S, T;

    binaryrot(BufferedReader in, PrintWriter out) throws IOException {
        StringTokenizer st = new StringTokenizer(in.readLine());
        N = Integer.parseInt(st.nextToken());
        S = new boolean[N];
        T = new boolean[N];
        int num1S = 0, num1T = 0;
        String str = in.readLine();
        for (int i = 0; i < N; i++) {
            if (str.charAt(i) == '1') {
                S[i] = true;
                num1S++;
            } else S[i] = false;
        }
        str = in.readLine();
        for (int i = 0; i < N; i++) {
            if (str.charAt(i) == '1') {
                T[i] = true;
                num1T++;
            } else T[i] = false;
        }

        if (num1S != num1T) {
            out.println(-1);
            return;
        }

        for (int i = 0; i < N; i++) {
            if (S[i] && T[i]) {
                S[i] = false;
                T[i] = false;
            }
        }

        int minStack = 0, maxStack = 0, currStack = 0;
        for (int i = 0; i < N; i++) {
            if (S[i]) currStack++;
            else if (T[i]) currStack--;
            minStack = Math.min(minStack, currStack);
            maxStack = Math.max(maxStack, currStack);
        }
        out.println(maxStack - minStack);
    }

    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        new binaryrot(in, out);
        in.close();
        out.close();
    }
}
