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;
const int N = 1e6 + 5;
int n;
string a, b;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> a >> b;
  string s;
  for (int i = 0; i < n; i++) {
    if (a[i] != b[i]) {
      s += a[i];
    }
  }
  if (((int)(s).size()) & 1) {
    cout << -1 << '\n';
    return 0;
  }
  if (!((int)(s).size())) {
    cout << 0 << '\n';
    return 0;
  }
  set<int> A, B;
  for (int i = 0; i < ((int)(s).size()); i++)
    if (s[i] == '0')
      A.insert(i);
    else
      B.insert(i);
  int ans = 0;
  while (((int)(A).size())) {
    ans++;
    set<int>& x = (*A.begin() < *B.begin()) ? A : B;
    set<int>& y = (*A.begin() < *B.begin()) ? B : A;
    int last = -1;
    while (((int)(x).size())) {
      auto itx = x.lower_bound(last);
      if (itx == x.end()) break;
      auto ity = y.lower_bound(*itx);
      if (ity == y.end()) break;
      last = *ity;
      x.erase(itx);
      y.erase(ity);
    }
  }
  cout << ans << '\n';
  return 0;
}
