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.
// package Quarantine;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class BinarySubsequenceRotation {
    public static void main(String[] args)throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int n=Integer.parseInt(br.readLine());
        char a[]=br.readLine().toCharArray();
        char b[]=br.readLine().toCharArray();
        int o=0,z=0;
        for(int i=0;i<n;i++){
            if(a[i]=='0'){
                z++;
            }
            else{
                o++;
            }
            if(b[i]=='0'){
                z--;
            }
            else{
                o--;
            }
        }
        if(o!=0||z!=0){
            System.out.println(-1);
            return;
        }
        int d[]=new int[n];
        for(int i=0;i<n;i++){
            if(a[i]!=b[i]){
                if(a[i]=='1'){
                    d[i]=1;
                }
                else{
                    d[i]=-1;
                }
            }
        }
        int op1=findMax(d,n);
        int op2=Math.abs(findMin(d,n));
        System.out.println(Math.max(op1,op2));
    }
    public static int findMax(int a[],int n){
        int max=0;
        int sum=0;
        for(int i=0;i<n;i++){
            if(a[i]+sum>=0){
                sum+=a[i];
            }
            else{
                sum=0;
            }
            max=Math.max(max,sum);
        }
        return max;
    }
    public static int findMin(int a[],int n){
        int min=0;
        int sum=0;
        for(int i=0;i<n;i++){
            if(a[i]+sum<=0){
                sum+=a[i];
            }
            else{
                sum=0;
            }
            min=Math.min(min,sum);
        }
        return min;
    }
}
