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 Codeforces
{
    public static void main(String args[])throws Exception
    {
        BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb=new StringBuilder();
        int n=Integer.parseInt(bu.readLine()),i;
        char s[]=bu.readLine().toCharArray();
        char t[]=bu.readLine().toCharArray();
        int a[]=new int[n],su=0;
        for(i=0;i<n;i++)
        {
            if(s[i]!=t[i])
            {
                if(s[i]=='1') a[i]=1;
                else a[i]=-1;
            }
            su+=a[i];
        }
        if(su!=0) {System.out.print("-1"); return;}
        int c1=0,c2=0,mx1=0,mx2=0;
        for(i=0;i<n;i++)
        {
            c1+=a[i]*1;
            mx1=Math.max(mx1,c1);
            if(c1<0) c1=0;
        }
        for(i=0;i<n;i++)
        {
            c2+=a[i]*-1;
            mx2=Math.max(mx2,c2);
            if(c2<0) c2=0;
        }
        System.out.print(Math.max(mx1,mx2));
    }
}