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;
int n;
string s, t;
void solve() {
  cin >> n >> s >> t;
  deque<int> d{0};
  int st = 0, n0 = 0, n1 = 0;
  for (int i = (0); i < (n); ++i)
    if (s[i] != t[i]) {
      if (s[i] == '0')
        ++n0;
      else
        ++n1;
      if (st == s[i] - '0') {
        ++d.back();
      } else {
        st ^= 1;
        d.push_back(1);
      }
    }
  if (n0 != n1) {
    cout << "-1\n";
    return;
  }
  if (d.front() == 0) d.pop_front();
  if (int((d).size()) & 1) {
    d.front() += d.back();
    d.pop_back();
  }
  int ans = 0;
  while (int((d).size())) {
    int m = *min_element((d).begin(), (d).end());
    for (auto& x : (d)) x -= m;
    ans += m;
    deque<int> dt = {0};
    for (int i = (0); i < (int((d).size())); ++i) {
      if (d[i])
        dt.push_back(d[i]);
      else if (++i < int((d).size()))
        dt.back() += d[i];
    }
    if (dt.front() == 0) dt.pop_front();
    d = dt;
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
}
