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.
#include <bits/stdc++.h>
using namespace std;
char s[1000010];
char t[1000010];
int n;
void solve() {
  cin >> n >> s >> t;
  char prv = 'a';
  int ans = 0;
  int c = 0;
  int par = 0;
  int n0 = 0, n1 = 0;
  for (int i = 0; i < 2 * n; ++i) {
    int j = i % n;
    if (s[j] == t[j]) continue;
    if (s[j] == '1') {
      ++par;
      if (n1 > 0)
        ++n1;
      else
        n1 = 1;
      --n0;
    } else {
      --par;
      if (n0 > 0)
        ++n0;
      else
        n0 = 1;
      --n1;
    }
    if (ans < n1) ans = n1;
    if (ans < n0) ans = n0;
  }
  if (par) ans = -1;
  cout << ans << endl;
}
int main() {
  solve();
  return 0;
}
